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Chapter 20

Trigonometrical Ratios

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 20(A)

Question 1

Look at the figures given below :

Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.
Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.
Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

From these figures, write down the values of :

(i) sin x

(ii) tan x

(iii) sec x

(iv) cos y

(v) cot y

(vi) cosec y

(vii) sin z

(viii) cos z

(ix) tan z

Answer

(i) sin x = perpendicularhypotenuse=qr\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{q}{r}

(ii) tan x = perpendicularbase=qp\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{q}{p}

(iii) sec x = hypotenusebase=rp\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{r}{p}

(iv) cos y = basehypotenuse=bn\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{b}{n}

(v) cot y = baseperpendicular=bm\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{b}{m}

(vi) cosec y = hypotenuseperpendicular=nm\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{n}{m}

(vii) sin z = perpendicularhypotenuse=un\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{u}{n}

(viii) cos z = basehypotenuse=kn\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{k}{n}

(ix) tan z = perpendicularbase=uk\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{u}{k}

Question 2

In the given figure, ∠B = 90°, AB = 4 units and BC = 3 units. Find:

In the given figure, ∠B = 90, AB = 4 units and BC = 3 units. Find. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

(i) sin A

(ii) cos A

(iii) cot A

(iv) sin C

(v) sec C

(vi) tan C

Answer

In triangle ABC,

By pythagoras theorem,

AC2 = AB2 + BC2

AC2 = 42 + 32

AC2 = 16 + 9

AC2 = 25

AC = 25\sqrt{25}

AC = 5 units

(i) sin A = perpendicularhypotenuse=BCAC=35\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3}{5}

(ii) cos A = basehypotenuse=ABAC=45\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4}{5}

(iii) cot A = baseperpendicular=ABBC=43\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{AB}{BC} = \dfrac{4}{3}

(iv) sin C = perpendicularhypotenuse=ABAC=45\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4}{5}

(v) sec C = hypotenusebase=ACBC=53\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{AC}{BC} = \dfrac{5}{3}

(vi) tan C = perpendicularbase=ABBC=43\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{BC} = \dfrac{4}{3}

Question 3

From the given figure, write down the values of :

From the given figure, write down the values of. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

(i) sin B

(ii) tan B

(iii) cos C

(iv) cot C

(v) (sin B cos C + cos B sin C)

(vi) (sec2 C - tan2 C)

Answer

Given a right triangle ABC with hypotenuse BC = 17 units and AB = 15 units.

First, find AC using the Pythagoras theorem :

BC2 = AB2 + AC2

AC2 = BC2 - AB2

AC2 = 172 - 152

AC2 = 289 - 225

AC2 = 64

AC = 64\sqrt{64}

AC = 8 units

(i) sin B = perpendicularhypotenuse=ACBC=817\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AC}{BC} = \dfrac{8}{17}

(ii) tan B = perpendicularbase=ACAB=815\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AC}{AB} = \dfrac{8}{15}

(iii) cos C = basehypotenuse=ACBC=817\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AC}{BC} = \dfrac{8}{17}

(iv) cot C = baseperpendicular=ACAB=815\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{AC}{AB} = \dfrac{8}{15}

(v) We have to find

sin B cos C + cos B sin C

First we will find the values of cos B & sin C

cos B = basehypotenuse=ABBC=1517\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AB}{BC} = \dfrac{15}{17}

sin C = perpendicularhypotenuse=ABBC=1517\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AB}{BC} = \dfrac{15}{17}

Substituting the values, we get :

sin B cos C + cos B sin C=817×817+1517×1517=64289+225289=289289=1.\text{sin B cos C + cos B sin C} = \dfrac{8}{17}\times \dfrac{8}{17} + \dfrac{15}{17} \times \dfrac{15}{17} \\[1em] = \dfrac{64}{289} + \dfrac{225}{289} \\[1em] = \dfrac{289}{289} \\[1em] = 1.

Hence, sin B cos C + cos B sin C = 1.

(vi) We have to find out

sec2C - tan2C

First we will find out the values of sec C & tan C

sec C = hypotenusebase=BCAC=178\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{BC}{AC} = \dfrac{17}{8}

tan C = perpendicularbase=ABAC=158\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{AC} = \dfrac{15}{8}

Now putting the values of sec C & tan C

sec2C - tan2C

= (178)2−(158)2\Big(\dfrac{17}{8}\Big)^2 - \Big(\dfrac{15}{8}\Big)^2

= 28964−22564=289−22564\dfrac{289}{64} - \dfrac{225}{64} = \dfrac{289 - 225}{64}

= 6464\dfrac{64}{64}

= 1.

Hence, sec2C - tan2C = 1.

Question 4

In the given figure, AD ⊥ BC.

In the given figure, AD. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

If AB = 13 cm, BD = 5 cm and DC = 16 cm, find the values of :

(i) sin B

(ii) sec B

(iii) cot B

(iv) cos C

(v) cosec C

(vi) tan C

Answer

Using pythagoras theorem in right angled triangle ADB

AB2 = BD2 + AD2

AD2 = AB2 - BD2

AD2 = 132 - 52

AD2 = 169 - 25

AD2 = 144

AD = 144\sqrt{144}

AD = 12 cm

Now we will find out AC using pythagoras theorem in right angled triangle ADC,

AC2 = DC2 + AD2

AC2 = 162 + 122

AC2 = 256 + 144

AC2 = 400

AC = 400\sqrt{400}

AC = 20 cm

Now,

(i) sin B = perpendicularhypotenuse=ADAB=1213\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AD}{AB} = \dfrac{12}{13}

(ii) sec B = hypotenusebase=ABBD=135\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{AB}{BD} = \dfrac{13}{5}

(iii) cot B = baseperpendicular=BDAD=512\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{BD}{AD} = \dfrac{5}{12}

(iv) cos C = basehypotenuse=DCAC=1620=45\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{DC}{AC} = \dfrac{16}{20} = \dfrac{4}{5}

(v) cosec C = hypotenuseperpendicular=ACAD=2012=53\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{AC}{AD} = \dfrac{20}{12} = \dfrac{5}{3}

(vi) tan C = perpendicularbase=ADDC=1216=34\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AD}{DC} = \dfrac{12}{16} = \dfrac{3}{4}

Question 5

If sin θ = 12\dfrac{1}{\sqrt{2}}, find the values of other trigonometrical ratios for θ.

Answer

sin θ = perpendicularhypotenuse=12\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}}

Let perpendicular = x and hypotenuse = 2x\sqrt{2}x

By using Pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (2x)2(\sqrt{2}x)^2 - x2

Base2 = 2x2 - x2

Base2 = x2

Base = x

Now, calculating the remaining trigonometric ratios :

cos θ = basehypotenuse=x2x=12\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{2}x} = \dfrac{1}{\sqrt{2}}

tan θ = perpendicularbase=xx=1\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{x}{x} = 1

cot θ = baseperpendicular=xx=1\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{x}{x} = 1

sec θ = hypotenusebase=2xx=2\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{\sqrt{2}x}{x} = \sqrt{2}

cosec θ = hypotenuseperpendicular=2xx=2\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{\sqrt{2}x}{x} = \sqrt{2}

Question 6

If tan θ = 815\dfrac{8}{15}, find the values of other trigonometrical ratios for θ.

Answer

tan θ = PerpendicularBase=815\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{8}{15}

Let perpendicular = 8x and base = 15x

By pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = (15x)2 + (8x)2

Hypotenuse2 = 225x2 + 64x2

Hypotenuse2 = 289x2

Hypotenuse = 289x2\sqrt{289x^2}

Hypotenuse = 17x

Now, calculating the remaining trigonometric ratios :

sin θ = PerpendicularHypotenuse=8x17x=817\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{8x}{17x} = \dfrac{8}{17}.

cos θ = BaseHypotenuse=15x17x=1517\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{15x}{17x} = \dfrac{15}{17}

cot θ = BasePerpendicular=15x8x=158\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{15x}{8x} = \dfrac{15}{8}

sec θ = HypotenuseBase=17x15x=1715\dfrac{\text{Hypotenuse}}{\text{Base}} = \dfrac{17x}{15x}= \dfrac{17}{15}

cosec θ = HypotenusePerpendicular=17x8x=178\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{17x}{8x} = \dfrac{17}{8}

Question 7

If cosec θ = 10\sqrt{10}, find the values of other trigonometrical ratios for θ.

Answer

cosec θ = hypotenuseperpendicular=101\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{\sqrt{10}}{1}.

Let hypotenuse = 10x\sqrt{10}x and perpendicular = x.

By pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (10x)2(\sqrt{10}x)^2 - x2

Base2 = 10x2 - x2

Base2 = 9x2

Base = (9x2)(\sqrt{9x^2})

Base = 3x

sin θ = perpendicularhypotenuse=x10x=110\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{x}{\sqrt{10}x} = \dfrac{1}{\sqrt{10}}

cos θ = basehypotenuse=3x10x=310\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{3x}{\sqrt{10}x} = \dfrac{3}{\sqrt{10}}

tan θ = perpendicularbase=x3x=13\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{x}{3x} = \dfrac{1}{3}

cot θ = baseperpendicular=3xx=3\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{3x}{x} = 3

sec θ = hypotenusebase=10x3x=103\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{\sqrt{10}x}{3x} = \dfrac{\sqrt{10}}{3}

Question 8

If sin θ = 35\dfrac{3}{5} and θ is an acute angle, find the values of cos θ and tan θ.

Answer

sin θ = perpendicularhypotenuse=35\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{3}{5}

Let perpendicular = 3x and hypotenuse = 5x

By using Pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (5x)2 - (3x)2

Base2 = 25x2 - 9x2

Base2 = 16x2

Base = 4x

cos θ = basehypotenuse=4x5x=45\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{4x}{5x} = \dfrac{4}{5}

tan θ = perpendicularbase=3x4x=34\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{3x}{4x} = \dfrac{3}{4}

Question 9

If tan θ = 512\dfrac{5}{12} and θ is acute, find the values of sin θ and cos θ.

Answer

tan θ = perpendicularbase=512\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{5}{12}

Let perpendicular = 5x and base = 12x

By pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = (12x)2 + (5x)2

Hypotenuse2 = 144x2 + 25x2

Hypotenuse2 = 169x2

Hypotenuse = 169x2\sqrt{169x^2}

Hypotenuse = 13x

sin θ = perpendicularhypotenuse=5x13x=513\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{5x}{13x} = \dfrac{5}{13}

cos θ = basehypotenuse=12x13x=1213\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{12x}{13x} = \dfrac{12}{13}

Question 10

If sin θ = 32\dfrac{\sqrt{3}}{2}, find the value of (cosec θ + cot θ).

Answer

sin θ = perpendicularhypotenuse=32\dfrac{\text{perpendicular}}{\text{hypotenuse}} =\dfrac{\sqrt{3}}{2}

Let perpendicular = 3x\sqrt{3}x and hypotenuse = 2x

We will find the value of base using pythagoras theorem

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (2x)2 - (3x)2(\sqrt{3}x)^2

Base2 = 4x2 - 3x2

Base2 = x2

Base = x

cosec θ = hypotenuseperpendicular=2x3x=23\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{2x}{\sqrt{3}x} = \dfrac{2}{\sqrt{3}}

cot θ = baseperpendicular=x3x=13\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{x}{\sqrt{3}x} = \dfrac{1}{\sqrt{3}}

Substituting above values in cosec θ + cot θ, we get :

cosec θ + cot θ = 23+13\dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}

= 33=3\dfrac{3}{\sqrt{3}} = {\sqrt{3}}.

Hence, cosec θ + cot θ = 3\sqrt{3}.

Question 11

If 13 sin θ = 5, find the value of

5 sin θ - 2 cos θtan θ\dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}}

Answer

13 sin θ = 5

sin θ = 513\dfrac{5}{13}

sin θ = perpendicularhypotenuse=513\dfrac{\text{perpendicular}}{\text{hypotenuse}} =\dfrac{\text{5}}{\text{13}}

Let perpendicular = 5x and hypotenuse = 13x

By pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (13x)2 - (5x)2

Base2 = 169x2 - 25x2

Base2 = 144x2

Base = 144x2\sqrt{144x^2}

Base = 12x

cos θ = basehypotenuse=12x13x=1213\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{12x}{13x} = \dfrac{12}{13}

tan θ = perpendicularbase=5x12x=512\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{5x}{12x} = \dfrac{5}{12}

Substituting values, we get :

⇒5 sin θ - 2 cos θtan θ=5×513−2×1213512=2513−2413512=113512=1265.\Rightarrow \dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \dfrac{5 \times \dfrac{5}{13} - 2 \times \dfrac{12}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{25}{13} - \dfrac{24}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{\dfrac{1}{13}}{\dfrac{5}{12}} \\[1em] = \dfrac{12}{65}.

Hence, 5 sin θ - 2 cos θtan θ=1265\dfrac{\text{5 sin θ - 2 cos θ}}{\text{tan θ}} = \dfrac{12}{65}.

Question 12

If cot θ = 13\dfrac{1}{\sqrt{3}}, show that (1−cos2θ2−sin2θ)=35\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}.

Answer

cot θ = baseperpendicular=13\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{1}{\sqrt{3}}

Let base = x and perpendicular = 3\sqrt{3}x

By using pythagoras theorem, we get :

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = (x)2 + (3x)2(\sqrt{3}x)^2

Hypotenuse2 = x2 + 3x2

Hypotenuse2 = 4x2

Hypotenuse = 4x2\sqrt{4x^2}

Hypotenuse = 2x

Now

sin θ = perpendicularhypotenuse=3x2x=32\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{\sqrt{3}x}{2x} = \dfrac{\sqrt{3}}{2}

cos θ = basehypotenuse=x2x=12\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{x}{2x} = \dfrac{1}{2}

Substituting values we get :

⇒(1−cos2θ2−sin2θ)=1−(12)22−(32)2=1−142−34=4−148−34=3454=35.\Rightarrow \Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{1 - \Big(\dfrac{1}{2}\Big)^2}{2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2} \\[1em] = \dfrac{1 - \dfrac{1}{4}}{2 - \dfrac{3}{4}} \\[1em] = \dfrac{\dfrac{4 - 1}{4}}{\dfrac{8 - 3}{4}} \\[1em] = \dfrac{\dfrac{3}{4}}{\dfrac{5}{4}} \\[1em] = \dfrac{3}{5}.

Hence, (1−cos2θ2−sin2θ)=35\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}.

Question 13

If sec θ = 135\dfrac{13}{5}, show that 2 sin θ - 3 cos θ4 sin θ - 9 cos θ\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3.

Answer

sec θ = hypotenusebase=135\dfrac{\text{hypotenuse}}{\text{base}} =\dfrac{13}{5}

Let hypotenuse = 13x and base = 5x

We will find perpendicular by using pythagoras theorem

Hypotenuse2 = Base2 + Perpendicular2

Perpendicular2 = Hypotenuse2 - Base2

Perpendicular2 = (13x)2 - (5x)2

Perpendicular2 = 169x2 - 25x2

Perpendicular2 = 144x2

Perpendicular = 144x2\sqrt{144x^2}

Perpendicular = 12x

Now

sin θ = perpendicularhypotenuse=12x13x=1213\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{12x}{13x} = \dfrac{12}{13}

cos θ = basehypotenuse=5x13x=513\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{5x}{13x} = \dfrac{5}{13}

Substituting values we get :

⇒2 sin θ - 3 cos θ4 sin θ - 9 cos θ=2×1213−3×5134×1213−9×513=2413−15134813−4513=913313=93=3.\Rightarrow \dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = \dfrac{2 \times \dfrac{12}{13} - 3 \times \dfrac{5}{13}}{4 \times \dfrac{12}{13} - 9 \times \dfrac{5}{13}} \\[1em] = \dfrac{\dfrac{24}{13} - \dfrac{15}{13}}{\dfrac{48}{13} - \dfrac{45}{13}} \\[1em] = \dfrac{\dfrac{9}{13}}{\dfrac{3}{13}} \\[1em] = \dfrac{9}{3} \\[1em] = 3.

Hence, proved that 2 sin θ - 3 cos θ4 sin θ - 9 cos θ=3\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3.

Question 14

If 3 tan θ = 4, show that (3sin⁡θ+2cos⁡θ3sin⁡θ−2cos⁡θ)\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)= 3.

Answer

tan θ = perpendicularbase=43\dfrac{\text{perpendicular}}{\text{base}}=\dfrac{4}{3}

Let Perpendicular = 4x and Base = 3x

We will find hypotenuse by using pythagoras theorem

Hypotenuse2 = Perpendicular2 + Base2

Hypotenuse2 = (4x)2 + (3x)2

Hypotenuse2 = 16x2 + 9x2

Hypotenuse2 = 25x2

Hypotenuse = 5x

Now

sin θ = perpendicularhypotenuse=4x5x=45\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}

cos θ = basehypotenuse=3x5x=35\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{3x}{5x} = \dfrac{3}{5}

Substituting values we get :

⇒3 sin θ + 2 cos θ3 sin θ - 2 cos θ=3×45+2×353×45−2×35=125+65125−65=12+6512−65=18565=18×55×6=186=3\Rightarrow \dfrac{\text{3 sin θ + 2 cos θ}}{\text{3 sin θ - 2 cos θ}} \\[1em] = \dfrac{3\times\dfrac{4}{5} + 2\times\dfrac{3}{5}}{3\times\dfrac{4}{5} -2\times\dfrac{3}{5}} \\[1em] = \dfrac{\dfrac{12}{5} + \dfrac{6}{5}}{\dfrac{12}{5} -\dfrac{6}{5}}\\[1em] = \dfrac{\dfrac{12+6}{5}}{\dfrac{12-6}{5}} \\[1em] = \dfrac{\dfrac{18}{5}}{\dfrac{6}{5}} \\[1em] = \dfrac{18\times5}{5\times6}\\[1em] = \dfrac{18}{6} = 3

Hence, proved that (3sin⁡θ+2cos⁡θ3sin⁡θ−2cos⁡θ)\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big) = 3.

Question 15

If cot θ = qp\dfrac{q}{p}, show that (psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ)=p2−q2p2+q2\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}.

Answer

cot θ = baseperpendicular=qp\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{q}{p}

Let base = qx and perpendicular = px

We will find hypotenuse by using pythagoras theorem

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = (qx)2 + (px)2

Hypotenuse2 = (q2 + p2)x2

Hypotenuse = (q2+p2)x2\sqrt{(q^2 + p^2)x^2}

Hypotenuse = (q2+p2)x\sqrt{(q^2 + p^2)}x

Now,

sin θ = perpendicularhypotenuse=px(p2+q2)x=pp2+q2\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{px}{\sqrt{(p^2 + q^2)}x} = \dfrac{p}{\sqrt{p^2 + q^2}}

cos θ = basehypotenuse=qx(p2+q2)x=qp2+q2\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{qx}{\sqrt{(p^2 + q^2)}x} = \dfrac{q}{\sqrt{p^2 + q^2}}

Substituting values we get :

⇒psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ=p×pp2+q2−q×qp2+q2p×pp2+q2+q×qp2+q2=p2p2+q2−q2p2+q2p2p2+q2+q2p2+q2=p2−q2p2+q2p2+q2p2+q2=(p2−q2)×p2+q2(p2+q2)×p2+q2=p2−q2p2+q2.\Rightarrow \dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\\[1em] = \dfrac{p\times\dfrac{p}{\sqrt{p^2 + q^2}} - q\times\dfrac{q}{\sqrt{p^2 + q^2}}}{p\times\dfrac{p}{\sqrt{p^2 + q^2}} + q\times\dfrac{q}{\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{\dfrac{p^2}{\sqrt{p^2 + q^2}} - \dfrac{q^2}{\sqrt{p^2 + q^2}} }{\dfrac{p^2}{\sqrt{p^2 + q^2}} + \dfrac{q^2}{\sqrt{p^2 + q^2}}}\\[1em] = \dfrac{\dfrac{p^2 - q^2}{\sqrt{p^2 + q^2}}}{\dfrac{p^2 + q^2}{\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{(p^2 - q^2)\times {\sqrt{p^2 + q^2}} }{(p^2 + q^2)\times {\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{p^2 - q^2}{p^2 + q^2}.

Hence, proved that psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ} = p2−q2p2+q2\dfrac{p^2 - q^2}{p^2 + q^2}.

Question 16

If 4 cot θ = 3, show that (sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)=17\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}.

Answer

cot θ = baseperpendicular=34\dfrac{\text{base}}{\text{perpendicular}}=\dfrac{3}{4}

Let base = 3x and perpendicular = 4x

We will find hypotenuse by using pythagoras theorem

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = (3x)2 + (4x)2

Hypotenuse2 = 9x2 + 16x2

Hypotenuse2 = 25x2

Hypotenuse = 5x

Now

sin θ = perpendicularhypotenuse=4x5x=45\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}

cos θ = basehypotenuse=3x5x=35\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{3x}{5x} = \dfrac{3}{5}

Substituting values we get :

⇒sin⁡θ−cos⁡θsin⁡θ+cos⁡θ=45−3545+35=4−354+35=1575=15×57=17.\Rightarrow \dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\\[1em] = \dfrac{\dfrac{4}{5} -\dfrac{3}{5}}{\dfrac{4}{5} + \dfrac{3}{5}}\\[1em] = \dfrac{\dfrac{4-3}{5}}{\dfrac{4+3}{5}}\\[1em] = \dfrac{\dfrac{1}{5}}{\dfrac{7}{5}}\\[1em] = \dfrac{1}{5}\times\dfrac{5}{7}\\[1em] = \dfrac{1}{7}.

Hence, proved that sin⁡θ−cos⁡θsin⁡θ+cos⁡θ\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ} = 17\dfrac{1}{7}

Question 17

Use the adjoining figure and write the values of :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

Use the adjoining figure and write the values of. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

In right angled triangle DBC,

Perpendicular = BC = 8 cm

Base = DB = 6 cm

Then we will find hypotenuse (CD) by pythagoras theorem,

Hypotenuse2 = Base2 + Perpendicular2

Hypotenuse2 = 62 + 82

Hypotenuse2 = 36 + 64

Hypotenuse2 = 100

Hypotenuse = 10 cm

In right angled triangle ABC,

Perpendicular = CB = 8 cm

Hypotenuse = AC = 17 cm

Let AD = m

Base (AB) = AD + DB = m + DB

By pythagoras theorem,

Base2 = Hypotenuse2 - Perpendicular2

(m + 6)2 = 172 - 82

m2 + 36 + 12m = 289 - 64

m2 + 36 + 12m = 225

m2 + 12m + 36 - 225 = 0

m2 + 12m - 189 = 0

m2 + 21m - 9m - 189 =0

m(m + 21) - 9(m + 21) = 0

(m + 21)(m - 9) = 0

m = -21 or m = 9

Sicne, length can't be negative.

so, m = 9 cm

AB = m + 6 = 9 + 6 = 15 cm

(i) sin x° = perpendicularhypotenuse=BCAC=817\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8}{17}.

(ii) cos y° = basehypotenuse=DBDC=610=35\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{DB}{DC} = \dfrac{6}{10} = \dfrac{3}{5}.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan x° = perpendicularbase=BCAB=815\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{BC}{AB} = \dfrac{8}{15}

sin y° = perpendicularhypotenuse=BCDC=810=45\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{BC}{DC} = \dfrac{8}{10} = \dfrac{4}{5}

Putting values of tan x°, sin y°, cos y° in 3 tan x° - 2 sin y° + 4 cos y°

= 3×815−2×45+4×353\times\dfrac{8}{15} - 2\times\dfrac{4}{5} + 4\times\dfrac{3}{5}

= 85−85+125\dfrac{8}{5} - \dfrac{8}{5} + \dfrac{12}{5}

= 125=225\dfrac{12}{5} = 2\dfrac{2}{5}.

Question 18

Using the adjoining figure, calculate the values of :

Using the adjoining figure, calculate the values of. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

(i) cos θ

(ii) tan Φ

(iii) cosec Φ

Answer

Using the adjoining figure, calculate the values of. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

In right angled triangle ABC,

Hypotenuse = AC = 13 units

Perpendicular = BC = 5 units

By pythagoras theorem,

Base2 = Hypotenuse2 - Perpendicular2

Base2 = 132 - 52

Base2 = 169 - 25

Base2 = 144

Base = 12 units

AB = 12 units.

Draw a perpendicular CE on AD.

In triangle CED,

CE = AB = 12 units and AE = BC = 5 units

From figure,

AD = DE + AE

DE = AD - AE = 14 - 5 = 9 units

In Triangle CED,

By pythagoras theorem,

CD2 = CE2 + ED2

CD2 = 122 + 92

CD2 = 144 + 81

CD2 = 225

CD = 225\sqrt{225} = 15 units.

(i) cos θ = basehypotenuse=ABAC=1213\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{AB}{AC} = \dfrac{12}{13}.

(ii) tan Φ = perpendicularbase=CEDE=129=43\dfrac{\text{perpendicular}}{\text{base}}= \dfrac{CE}{DE} = \dfrac{12}{9} = \dfrac{4}{3}.

(iii) cosec Φ = hypotenuseperpendicular=CDCE=1512=54\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{CD}{CE} = \dfrac{15}{12} = \dfrac{5}{4}.

Question 19

If (tan θ + cot θ) = 5, find the value of (tan2θ + cot2θ).

Answer

As, (tan θ + cot θ) = 5

Squaring both sides, we get :

⇒ (tan θ + cot θ)2 = 52

⇒ tan2θ + cot2θ + 2 tan θ cot θ = 25

⇒ tan2θ + cot2θ + 2tan⁡θ×1tan⁡θ2\tan \theta \times \dfrac{1}{\tan \theta} = 25

⇒ tan2θ + cot2θ + 2 = 25

⇒ tan2θ + cot2θ = 25 - 2

⇒ tan2θ + cot2θ = 23.

Hence, tan2θ + cot2θ = 23.

Question 20

If (cos θ + sec θ ) = 52\dfrac{5}{2}, find the value of (cos2θ + sec2θ).

Answer

Given,

(cos θ + sec θ ) = 52\dfrac{5}{2}

Squaring both sides,

cos2θ + sec2θ + 2cos θ sec θ = 254\dfrac{25}{4}

As we know cos θ = 1sec⁡θ\dfrac{1}{\sec θ }

cos2θ + sec2θ + 2 = 254\dfrac{25}{4}

⇒cos⁡2θ+sec⁡2θ=254−2\Rightarrow \cos^2θ + \sec^2θ = \dfrac{25}{4} - 2

⇒25−84=174\Rightarrow \dfrac{25 - 8}{4} = \dfrac{17}{4}

Hence, cos2θ + sec2θ = 174\dfrac{17}{4}.

Question 21

Evaluate x and y from the given figure.

Evaluate x and y from the given figure. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

In the given figure there are two right angled triangles, △ADC and △BDC.

In △ADC,

∠ACD = 60° and AC = 10 m, CD = x m

cos 60° = basehypotenuse\dfrac{\text{base}}{\text{hypotenuse}}

12=CDAC\dfrac{1}{2} = \dfrac{CD}{AC}

12=x10\dfrac{1}{2} = \dfrac{x}{10}

x = 5 m.

In △BDC,

BC = 525\sqrt{2} m

CD = x = 5 m

sin y° = perpendicularhypotenuse=DCBC=552=12\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{DC}{BC} = \dfrac{5}{5\sqrt{2}} = \dfrac{1}{\sqrt{2}}

sin y° = 12\dfrac{1}{\sqrt{2}}

sin y° = sin 45°

y° = 45°.

Hence, x = 5 m and y° = 45°.

Question 22

In the given figure, △ABC is right angled at B.

If AC = 20 cm and tan A = 34\dfrac{3}{4}, find the lengths of AB and BC.

In the given figure, △ABC is right angled at B. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

tan A = perpendicularbase=BCAB\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{BC}{AB}

Given,

tan A = 34\dfrac{3}{4}

Let BC = 3x and AB = 4x.

Now by pythagoras theorem

AC2 = BC2 + AB2

(20)2 = (3x)2 + (4x)2

400 = 9x2 + 16x2

25x2 = 400

x2 = 16

x = 16\sqrt{16} = 4

AB = 4x = 16 cm and BC = 3x = 12 cm.

Hence, length of AB = 16 cm and BC = 12 cm.

Question 23

If cos θ = 2x1+x2\dfrac{2x}{1 + x^2}, find the values of sin θ and tan θ in terms of x.

find the values of sin θ and tan θ in terms of x. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

cos θ = basehypotenuse=2x1+x2\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{2x}{1 + x^2}

Let base = 2x and hypotenuse = 1 + x2

Now we will find perpendicular by using pythagoras theorem

Perpendicular2 = Hypotenuse2 - Base2

Perpendicular2 = (1 + x2)2 - (2x)2

Perpendicular2 = 1 + x4 + 2x2 - 4x2

Perpendicular2 = 1 + x4 - 2x2

Perpendicular2 = (x2 - 1)2

Perpendicular = (x2 - 1)

Now,

sin θ = perpendicularhypotenuse=x2−11+x2\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{x^2 - 1}{1 + x^2}

tan θ = perpendicularbase=x2−12x\dfrac{\text{perpendicular}}{\text {base}} = \dfrac{x^2 - 1}{2x}

Exercise 20(B)

Question 1

Without using trigonometric table, find the values of:

(i) sin 60° cos 30° + cos 60° sin 30°

(ii) sin 45° cos 30° - cos 45° sin 30°

(iii) cos 60° cos 45° + sin 60° sin 45°

(iv) cos 90° + cos2 45° sin 30° tan 45°

Answer

(i) sin 60° cos 30° + cos 60° sin 30°

= 32×32+12×12\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}

= 34+14\dfrac{3}{4} + \dfrac{1}{4}

= 3+14=44\dfrac{3+1}{4} = \dfrac{4}{4}

= 1.

Hence, sin 60° cos 30° + cos 60° sin 30° = 1.

(ii) sin 45° cos 30° - cos 45° sin 30°

= 12×32−12×12\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \times\dfrac{1}{2}

= 322−122\dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}

= 3−122\dfrac{\sqrt{3}-1}{2\sqrt{2}}.

Hence, sin 45° cos 30° - cos 45° sin 30° = 3−122\dfrac{\sqrt{3}-1}{2\sqrt{2}}.

(iii) cos 60° cos 45° + sin 60° sin 45°

= 12×12+32×12\dfrac{1}{2}\times\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}\times\dfrac{1}{\sqrt2}

= 122+322\dfrac{1}{2\sqrt{2}} + \dfrac{\sqrt{3}}{2\sqrt{2}}

= 1+322\dfrac{1 + \sqrt{3}}{2\sqrt{2}}.

Hence, cos 60° cos 45° + sin 60° sin 45° = 1+322\dfrac{1 + \sqrt{3}}{2\sqrt{2}}.

(iv) cos 90° + cos2 45° sin 30° tan 45°

As, cos2 45° = (cos 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

Therefore,

cos 90° + cos2 45° sin 30° tan 45°

= 0 + 12×12×1\dfrac{1}{2}\times\dfrac{1}{2}\times1

= 14\dfrac{1}{4}.

Hence, cos 90° + cos2 45° sin 30° tan 45° = 14\dfrac{1}{4}.

Question 2

Without using trigonometric tables, find the values of;

(i) (3sin2 45° + 2cos260°)

(ii) (3cos2 30° + tan260°)

(iii) (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

(iv) 2 2\sqrt{2} cos 45°cos 60° + 2 3\sqrt{3} sin 30° tan 60° - cos 0°

(v) 43\dfrac{4}{3} tan230°+ sin260° - 3 cos260°+ 34\dfrac{3}{4} tan260°- 2 tan245°

(vi) sin⁡245∘+cos⁡245∘tan⁡260∘\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ}

Answer

(i) As,

sin2 45° = (sin 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

and

cos2 60° = (cos 60°)2 = (12)2=14\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}

Substituting values we get :

(3sin2 45° + 2cos260°)

= 3×12+2×143\times\dfrac{1}{2} + 2\times\dfrac{1}{4}

= 32+24=32+12\dfrac{3}{2} + \dfrac{2}{4} = \dfrac{3}{2} + \dfrac{1}{2}

= 3+12=42\dfrac{3+1}{2} = \dfrac{4}{2} = 2.

Hence, 3 sin2 45° + 2 cos260° = 2.

(ii) As,

cos2 30° = (cos 30°)2 = (32)2=34\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}

and

tan2 60° = (tan 60°)2 = (3)2(\sqrt{3})^2 = 3

Therefore,

3cos2 30° + tan260°

= 3×343\times \dfrac{3}{4} + 3

= 94\dfrac{9}{4} + 3

= 9+124=214=514\dfrac{9 + 12}{4} = \dfrac{21}{4} = 5\dfrac{1}{4}.

Hence, 3cos2 30° + tan260° = 5145\dfrac{1}{4}.

(iii) (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

=(1+12+12)(1−12+12)=(22+2+222)(22−2+222)=(32+222)(32−222)=((32)2−(2)222×22)=(18−48)=148=74=134.= \Big( 1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \Big)\Big(1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big)\\[1em] = \Big(\dfrac{2\sqrt{2}+ 2 + \sqrt{2}}{2\sqrt{2}}\Big)(\dfrac{2\sqrt{2} - 2 + \sqrt{2}}{2\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{3\sqrt{2}+ 2}{2\sqrt{2}}\Big)\Big(\dfrac{3\sqrt{2}- 2}{2\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{(3\sqrt{2})^2- (2)^2 }{2\sqrt{2}\times {2\sqrt{2}}}\Big)\\[1em] = \Big(\dfrac{18-4}{8}\Big) \\[1em] = \dfrac{14}{8} \\[1em] = \dfrac{7}{4} \\[1em] = 1\dfrac{3}{4}.

Hence, (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°) = 1341\dfrac{3}{4}.

(iv) 222\sqrt{2} cos 45°cos 60° + 2 3\sqrt{3} sin 30° tan 60° - cos 0°

= 2 2×12×12+23×12×3−1\sqrt{2}\times\dfrac{1}{\sqrt{2}}\times\dfrac{1}{2} + 2\sqrt{3}\times\dfrac{1}{2}\times\sqrt{3} - 1

= 1 + 3 - 1

= 3.

Hence, 2 2\sqrt{2} cos 45°cos 60° + 2 3\sqrt{3} sin 30° tan 60° - cos 0° = 3.

(v) As,

cos2 60° = (cos 60°)2 = (12)2=14\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}

and

tan2 30° = (tan 30°)2 = (13)2=13\Big(\dfrac{1}{\sqrt{3}}\Big)^2 = \dfrac{1}{3}

tan2 45° = (tan 45°)2 = 1

tan2 60° = (tan 60°)2 = (3\sqrt{3})2 = 3

sin2 60° = (sin 60°)2 = (32)2=34\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}

Therefore,

43\dfrac{4}{3} tan230°+ sin260° - 3 cos260°+ 34\dfrac{3}{4} tan260°- 2 tan245°

43×13+34−3×14+34×3−2×1=49+34−34+94−2=49+94−2=16+81−7236=2536.\dfrac{4}{3}\times\dfrac{1}{3}+ \dfrac{3}{4} - 3 \times\dfrac{1}{4} + \dfrac{3}{4}\times 3- 2 \times 1 \\[1em] = \dfrac{4}{9}+ \dfrac{3}{4} -\dfrac{3}{4} + \dfrac{9}{4}- 2 \\[1em] = \dfrac{4}{9} + \dfrac{9}{4} - 2 \\[1em] = \dfrac{16 + 81 - 72}{36}\\[1em] = \dfrac{25}{36}.

Hence, 43\dfrac{4}{3} tan230°+ sin260° - 3 cos260°+ 34\dfrac{3}{4} tan260°- 2 tan245° = 2536\dfrac{25}{36}.

(vi) As,

cos2 45° = (cos 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

sin2 45° = (sin 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

tan2 60° = (tan 60°)2 = (3\sqrt{3})2 = 3

Therefore,

sin⁡245∘+cos⁡245∘tan⁡260∘\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ}

= 12+123\dfrac{\dfrac{1}{2}+\dfrac{1}{2}}{3}

= 13\dfrac{1}{3}

Hence, sin⁡245∘+cos⁡245∘tan⁡260∘=13\dfrac{\sin^2 45^\circ + \cos^2 45^\circ}{\tan^2 60^\circ} = \dfrac{1}{3}.

Question 3

Without using trigonometric tables, find the values of;

(i) sin⁡30∘−sin⁡90∘+2×cos⁡0∘tan⁡30∘×tan⁡60∘\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}

(ii) 5sin⁡230∘+cos⁡245∘−4tan⁡230∘2sin⁡30∘cos⁡30∘+tan⁡45∘\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}

(iii) tan⁡260∘+4cos⁡245∘+sec⁡230∘+5cos⁡290∘cosec⁡30∘+sec⁡60∘−cot⁡230∘\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}

(iv) 4(sin4 30° + cos4 60°) - 3 (cos2 45° - sin2 90°)

Answer

(i) Solving,

sin⁡30∘−sin⁡90∘+2×cos⁡0∘tan⁡30∘×tan⁡60∘=12−1+2×113×3=12−1+213×3=12+11=32.\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}\\[1em] = \dfrac{\dfrac{1}{2} - 1 +2\times1}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} - 1 + 2}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} + 1}{1}\\[1em] = \dfrac{3}{2}.

Hence, the required value = 32\dfrac{3}{2}.

(ii) As,

sin2 30° = (sin 30°)2 = (12)2=14\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}

cos2 45° = (cos 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

tan2 30° = (tan 30°)2 = (13)2=13\Big(\dfrac{1}{\sqrt{3}}\Big)^2 = \dfrac{1}{3}

Therefore,

5sin⁡230∘+cos⁡245∘−4tan⁡230∘2sin⁡30∘cos⁡30∘+tan⁡45∘=5×14+12−4×132×12×32+1=54+12−4332+1=15+6−161232+1=5123+22=56(2+3).\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}\\[1em] =\dfrac{5\times\dfrac{1}{4} + \dfrac{1}{2} - 4\times\dfrac{1}{3}}{2\times\dfrac{1}{2} \times\dfrac{\sqrt{3}}{2}+ 1 }\\[1em] = \dfrac{\dfrac{5}{4} + \dfrac{1}{2} -\dfrac{4}{3}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{15 + 6 - 16}{12}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{5}{12}}{\dfrac{\sqrt{3} + 2}{2}} \\[1em] = \dfrac{5}{6(2 + \sqrt{3})}.

Hence, 5sin⁡230∘+cos⁡245∘−4tan⁡230∘2sin⁡30∘cos⁡30∘+tan⁡45∘=56(2+3)\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ} = \dfrac{5}{6(2 + \sqrt{3})}.

(iii) As,

tan2 60° = (tan 60°)2 = (3\sqrt{3})2 = 3

cos2 45° = (cos 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

sec2 30° = (sec 30°)2 = (23)2=43\Big(\dfrac{2}{\sqrt{3}}\Big)^2 = \dfrac{4}{3}

cos2 90° = (cos 90°)2 = 0

cot2 30° = (cot 30°)2 = (3\sqrt{3})2 = 3

Therefore,

tan⁡260∘+4cos⁡245∘+sec⁡230∘+5cos⁡290∘cosec⁡30∘+sec⁡60∘−cot⁡230∘=3+4×12+43+02+2−3=3+2+43+01=5+43=193=613.\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}\\[1em] = \dfrac{3 + 4\times\dfrac{1}{2} + \dfrac{4}{3} + 0}{2 + 2 - 3}\\[1em] = \dfrac{3 + 2 + \dfrac{4}{3} + 0}{1}\\[1em] = 5+ \dfrac{4}{3}\\[1em] = \dfrac{19}{3} \\[1em] = 6\dfrac{1}{3}.

Hence, tan⁡260∘+4cos⁡245∘+sec⁡230∘+5cos⁡290∘cosec⁡30∘+sec⁡60∘−cot⁡230∘=193\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ} = \dfrac{19}{3}

(iv) As,

sin4 30° = (sin 30°)4 = (12)4=116\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}

cos4 60° = (cos 60°)4 = (12)4=116\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}

cos2 45° = (cos 45°)2 = (12)2=12\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}

sin2 90° = (sin 90°)2 = 1

Therefore,

4 (sin4 30° + cos4 60°) - 3 (cos2 45° - sin2 90°)

= 4(116+116)−3(12−1)4\Big(\dfrac{1}{16} + \dfrac{1}{16}\Big) - 3 \Big(\dfrac{1}{2} - 1\Big)

= 4(216)−3(−12)4\Big(\dfrac{2}{16}\Big) - 3 \Big(\dfrac{-1}{2}\Big)

= 12+32\dfrac{1}{2} + \dfrac{3}{2}

= 2.

Hence, 4 (sin4 30° + cos4 60°) - 3 (cos2 45° - sin2 90°) = 2.

Question 4

Verify each of the following :

(i) cos 60° cos 30° - sin 60° sin 30° = 0

(ii) cos 60° = (1 - 2 sin230°) = (2 cos230° - 1)

(iii) tan 30° = (tan⁡60∘−tan⁡30∘1+tan⁡60∘tan⁡30∘)\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)

Answer

(i) cos 60° cos 30° - sin 60° sin 30°

= 12×32−32×12\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}

= 0.

Hence, proved that cos 60° cos 30° - sin 60° sin 30° = 0.

(ii) As,

sin2 30° = (sin 30°)2 = (12)2=14\Big(\dfrac{1}{2}\Big)^2= \dfrac{1}{4}

cos2 30° = (cos 30°)2 = (32)2=34\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}

Therefore

Left Hand Side :

cos 60° = 12\dfrac{1}{2}

Right Hand Side :

(1 - 2 sin230°)

=1 - 2 ×14=12\times \dfrac{1}{4} = \dfrac{1}{2}

(2 cos230° - 1)

= 2 ×34\times\dfrac{3}{4} - 1 = 12\dfrac{1}{2}

Hence proved that cos 60° = (1 - 2 sin230°) = (2 cos230° - 1).

(iii) Left Hand Side :

tan 30° = 13\dfrac{1}{\sqrt{3}}

Right Hand Side

(tan⁡60∘−tan⁡30∘1+tan⁡60∘tan⁡30∘)=3−131+3×13=3−131+1=232=13.\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big) = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.

Hence proved that tan 30° = (tan⁡60∘−tan⁡30∘1+tan⁡60∘tan⁡30∘)\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)

Question 5

Verify each of the following :

(i) sin 60° cos 30° - cos 60° sin 30° = sin 30°

(ii) 2 sin 30° cos 30° = sin 60°

(iii) 2 sin 45° cos 45° = sin 90°

Answer

(i) Left Hand Side :

sin 60° cos 30° - cos 60° sin 30°

= 32×32−12×12\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} -\dfrac{1}{2}\times \dfrac{1}{2}

= 34−14=24\dfrac{3}{4} -\dfrac{1}{4} = \dfrac{2}{4}

= 12\dfrac{1}{2}.

Right Hand Side :

sin 30° = 12\dfrac{1}{2}

Hence, proved that sin 60° cos 30° - cos 60° sin 30° = sin 30°.

(ii) Left Hand Side :

2 sin 30° cos 30°

= 2×12×322\times\dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}

= 32\dfrac{\sqrt{3}}{2}

Right Hand Side :

sin 60° = 32\dfrac{\sqrt{3}}{2}

Hence, proved that 2 sin 30° cos 30° = sin 60°.

(iii) Left Hand Side :

2 sin 45° cos 45°

= 2×12×122\times\dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2}

= 1

Right Hand Side :

sin 90° = 1

Hence, proved that 2 sin 45° cos 45° = sin 90°.

Question 6

If A = 45°, verify that :

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = (2 cos2A - 1) = (1 - 2 sin2A)

Answer

(i) L.H.S. :

sin 2A = sin 2(45°) = sin 90°

= 1

R.H.S. :

2 sin A cos A = 2 sin 45° cos 45°

= 2×12×122\times\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}

= 1.

Hence, proved that sin 2A = 2 sin A cos A.

(ii) Substituting value of A = 45° in cos 2A, we get :

cos 2A = cos 2(45°)

= cos 90°

= 0.

Substituting value of A = 45° in 2 cos2A - 1, we get :

⇒ 2 cos2A - 1

= 2 cos245° - 1

= 2 (cos 45°)2 - 1

= 2(12)22\Big(\dfrac{1}{\sqrt{2}}\Big)^2 - 1

= 2×12−12 \times \dfrac{1}{2} - 1

= 1 - 1

= 0.

Substituting value of A = 45° in 1 - 2 sin2A, we get :

⇒ 1 - 2 sin2A

= 1 - 2 sin245°

= 1 - 2 (sin 45°)2

= 1 - 2(12)22\Big(\dfrac{1}{\sqrt{2}}\Big)^2

= 1 - 2×122 \times \dfrac{1}{2}

= 1 - 1

= 0.

Hence, proved that cos 2A = (2 cos2A - 1) = (1 - 2 sin2A).

Question 7

If A = 30°, prove that :

(i) sin 2A = 2tan⁡A1+tan⁡2A\dfrac{2 \tan A}{1 + \tan^2A}

(ii) cos 2A = (1−tan⁡2A1+tan⁡2A)\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big)

Answer

(i) Left Hand Side:

sin 2A = sin 2(30°) = sin 60°

= 32\dfrac{\sqrt{3}}{2}

Right Hand Side:

2tan⁡A1+tan⁡2A=2×131+13=2343=23×34=32.\dfrac{2 \tan A}{1 + \tan^2A} = \dfrac{2\times \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\\[1em] = \dfrac{2}{\sqrt{3}}\times \dfrac{3}{4}\\[1em] = \dfrac{\sqrt{3}}{2}.

Hence, proved that sin 2A = 2tan⁡A1+tan⁡2A\dfrac{2 \tan A}{1 + \tan^2A}

(ii) Left Hand Side :

cos 2A = cos 2(30°) = cos 60°

= 12\dfrac{1}{2}

Right Hand Side :

1−tan⁡2A1+tan⁡2A=1−tan⁡230°1+tan⁡230°=1−(13)21+(13)2=1−131+13=2343=12.\dfrac{1 - \tan^2A}{1 + \tan^2A} \\[1em] = \dfrac{1 - \tan^230°}{1 + \tan^230°} \\[1em] =\dfrac{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{1 - \dfrac{1}{3}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{1}{2}.

Hence, proved that cos 2A = (1−tan⁡2A1+tan⁡2A)\Big(\dfrac{1 - \tan^2A}{1 + \tan^2A}\Big).

Question 8

If A = B = 45°, show that :

(i) sin(A - B) = sin A cos B - cos A sin B

(ii) cos(A + B) = cosA cosB - sin A sin B

Answer

(i) Left Hand Side :

sin(A - B) = sin (45° - 45°) = sin 0

= 0

Right Hand Side :

sin A cos B - cos A sin B

= sin 45° cos 45° - cos 45° sin 45°

= 12×12−12×12\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}

= 0.

Hence, proved that sin(A - B) = sin A cos B - cos A sin B.

(ii) Left Hand Side :

cos(A + B) = cos (45° + 45°) = cos 90°

= 0.

Right Hand Side :

cos A cos B - sin A sin B

= cos 45° cos 45° - sin 45° sin 45°

= 12×12−12×12\dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\times\dfrac{1}{\sqrt{2}}

= 12−12\dfrac{1}{2} - \dfrac{1}{2}

= 0.

Hence, proved that cos(A + B) = cosA cosB - sin A sin B.

Question 9

If A = 60° and B = 30°, show that :

(sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = 1

Answer

Left Hand Side :

(sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2

= (sin 60° cos 30° + cos 60° sin 30°)2 + (cos 60° cos 30° - sin 60° sin 30°)2

= (32×32+12×12)2+(12×32−32×12)2\Big(\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\Big)^2

= (34+14)2+(34−34)2\Big(\dfrac{3}{4} + \dfrac{1}{4})^2 + \Big(\dfrac{\sqrt{3}}{4}- \dfrac{\sqrt{3}}{4}\Big)^2

= 1 + 0 = 1.

Right Hand Side = 1

Hence, proved that (sin A cos B + cos A sin B)2 + (cos A cos B - sin A sin B)2 = 1.

Question 10

If A = 60° and B = 30°, prove that :

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

(iii) cos (A - B) = cos A cos B + sin A sin B

(iv) tan (A - B) = tan⁡A−tan⁡B1+tan⁡Atan⁡B\dfrac{\tan A - \tan B}{1 + \tan A \tan B}

Answer

(i) Left Hand Side

sin (A + B) = sin (60° + 30°) = sin 90°

= 1

Right Hand Side

sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

= 32×32+12×12\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\times\dfrac{1}{2}

= 34+14\dfrac{3}{4} + \dfrac{1}{4}

= 1

Hence, proved that sin (A + B) = sin A cos B + cos A sin B.

(ii) Left Hand Side

cos (A + B) = cos (60° + 30°) = cos 90°

= 0

Right Hand Side

cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

= 12×32−32×12\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}

= 34−34\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}

= 0.

Hence, proved that cos (A + B) = cos A cos B - sin A sin B.

(iii) Left Hand Side

cos (A - B) = cos (60° - 30°) = cos 30°

= 32\dfrac{\sqrt{3}}{2}

Right Hand Side

cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

12×32+32×12=34+34=234=32\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}\\[1em] = \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4}\\[1em] = \dfrac{2\sqrt{3}}{4}\\[1em] =\dfrac{\sqrt{3}}{2}

Hence, proved that cos (A - B) = cos A cos B + sin A sin B.

(iv) Left Hand Side :

tan (A - B) = tan (60° - 30°) = tan 30°

= 13\dfrac{1}{\sqrt{3}}

Right Hand Side

⇒(tan⁡A−tan⁡B1+tan⁡Atan⁡B)=(tan⁡60∘−tan⁡30∘1+tan⁡60∘tan⁡30∘)=3−131+3×13=3−131+1=232=13.\Rightarrow \Big(\dfrac{\tan A - \tan B}{1 + \tan A \tan B}\Big)\\[1em] = \Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)\\[1em] = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.

Hence, proved that tan (A - B) = tan⁡A−tan⁡B1+tan⁡Atan⁡B\dfrac{\tan A - \tan B}{1 + \tan A \tan B}.

Question 11

Evaluate : cos⁡3A+2cos⁡4Asin⁡3A+2sin⁡4A\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}, when A = 15°.

Answer

Solving,

⇒cos⁡3A+2cos⁡4Asin⁡3A+2sin⁡4A=cos⁡3(15∘)+2cos⁡4(15∘)sin⁡3(15∘)+2sin⁡4(15∘)=cos⁡45∘+2cos⁡60∘sin⁡45∘+2sin⁡60∘=12+2×1212+2×32=12+112+3=1+221+62=1+21+6.\Rightarrow \dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A}\\[1em] = \dfrac{\cos 3(15^\circ) + 2 \cos 4(15^\circ)}{\sin 3(15^\circ) + 2\sin 4(15^\circ)}\\[1em] = \dfrac{\cos 45^\circ + 2 \cos 60^\circ}{\sin 45^\circ + 2\sin 60^\circ}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}+ 2\times\dfrac{\sqrt{3}}{2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}+ 1 }{\dfrac{1}{\sqrt{2}}+ {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1 + \sqrt{2}}{\sqrt{2}}}{\dfrac{1 + \sqrt{6}}{\sqrt{2}}}\\[1em] = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}.

Hence, cos⁡3A+2cos⁡4Asin⁡3A+2sin⁡4A=1+21+6\dfrac{\cos 3A + 2 \cos 4A}{\sin 3A + 2\sin 4A} = \dfrac{1 + \sqrt{2}}{1 + \sqrt{6}}.

Question 12

Evaluate : 3sin⁡3A+2cos⁡(2A+5∘)2cos⁡3A−sin⁡(2A−10∘)\dfrac{3\sin 3A + 2\cos (2A + 5^\circ)}{2\cos 3A - \sin (2A - 10^\circ)} , when A = 20°.

Answer

Solving,

⇒3sin⁡3(20∘)+2cos⁡(2(20∘)+5∘)2cos⁡3(20∘)−sin⁡(2(20∘)−10∘)=3sin⁡60∘+2cos⁡45∘2cos⁡60∘−sin⁡30∘=3×32+2×122×12−12=332+2212=332+212=33+222×21=33+22.\Rightarrow \dfrac{3\sin 3(20^\circ) + 2 \cos (2(20^\circ)+ 5^\circ)}{2 \cos 3(20^\circ) - \sin (2(20^\circ) - 10^\circ)}\\[1em] = \dfrac{3 \sin 60^\circ + 2 \cos 45^\circ}{2 \cos 60^\circ - \sin 30^\circ}\\[1em] = \dfrac{3\times\dfrac{\sqrt{3}}{2} + 2\times\dfrac{1}{\sqrt{2}}}{2\times\dfrac{1}{2} - \dfrac{1}{2}} \\[1em] = \dfrac{\dfrac{3\sqrt{3}}{2} + \dfrac{2}{\sqrt{2}}}{\dfrac{1}{{2}}}\\[1em] = \dfrac{\dfrac{3\sqrt{3}}{2} + {\sqrt{2}}}{\dfrac{1}{2}}\\[1em] = \dfrac{3\sqrt{3}+2\sqrt{2}}{2}\times\dfrac{2}{1}\\[1em] = {3\sqrt{3} + 2\sqrt{2}}.

Hence, 3sin⁡3A+2cos⁡(2A+5∘)2cos⁡3A−sin⁡(2A−10∘)=33+22.\dfrac{3\sin 3A + 2\cos (2A + 5^\circ)}{2\cos 3A - \sin (2A - 10^\circ)} = {3\sqrt{3} + 2\sqrt{2}}.

Question 13

Show that 4(sin430° + cos460°) - 3(cos245° - sin290°) = 2.

Answer

sin 30° = 12\dfrac{1}{2}

sin430° = (sin 30°)4 = 116\dfrac{1}{16}

cos 60° = 12\dfrac{1}{2}

cos460° = (cos 60°)4 = 116\dfrac{1}{16}

sin 45° = 12\dfrac{1}{\sqrt{2}}

cos245° = (cos 45°)2 = 12\dfrac{1}{2}

sin290° = (sin 90°)2 = 1

Left Hand Side

4(sin430° + cos460°)- 3(cos245°- sin290°)

= 4(116+116)−3(12−1)4\Big(\dfrac{1}{16}+\dfrac{1}{16}\Big) - 3 \Big(\dfrac{1}{2}- 1\Big)

= 4(216)−3(−12)4\Big(\dfrac{2}{16}\Big) - 3 \Big(\dfrac{-1}{2}\Big)

= 12+32\dfrac{1}{2} + \dfrac{3}{2}

= 2.

Right Hand Side = 2

Hence, proved that 4(sin430° + cos460°) - 3(cos245°- sin290°) = 2.

Question 14

Without using tables, verify that :

(i) cos 60° = (cos230° - sin230)

(ii) sin 60° = 2tan⁡30∘1+tan⁡230∘=32\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{\sqrt{3}}{2}

(iii) cos 60° = 1−tan⁡230∘1+tan⁡230∘=12\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{1}{2}

Answer

(i) cos230° = (cos 30°)2 = (32)2=34\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}

sin230° = (sin 30°)2 = (12)2=14\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}

Left Hand Side

cos 60° = 12\dfrac{1}{2}

Right Hand Side

cos230° - sin230°

= 34−14=12\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}

Hence, proved that cos 60° = (cos230° - sin230).

(ii) Left Hand Side :

sin 60° = 32\dfrac{\sqrt{3}}{2}

Right Hand Side :

tan2 30° = (tan 30°)2 = (13)2=13\Big(\dfrac{1}{\sqrt{3}}\Big)^2 = \dfrac{1}{3}

⇒2tan⁡30∘1+tan⁡230∘=2×131+13=2343=23×34=32.\Rightarrow \dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\dfrac{2\times\dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}\\[1em] = \dfrac{2}{\sqrt{3}}\times\dfrac{3}{4}\\[1em] = \dfrac{\sqrt{3}}{2}.

Hence, proved that sin 60° = 2tan⁡30∘1+tan⁡230∘=32\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{\sqrt{3}}{2}.

(iii) Left Hand Side

cos 60° = 12\dfrac{1}{2}

Right Hand Side

1−tan⁡230∘1+tan⁡230∘=1−131+13=3−133+13=24=12.\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ}\\[1em] =\dfrac{1 -\dfrac{1}{3}}{1 + \dfrac{1}{{3}}}\\[1em] = \dfrac{\dfrac{{3} - 1}{{3}}}{\dfrac{{3} + 1}{{3}}}\\[1em] =\dfrac{2}{4} \\[1em] = \dfrac{1}{2}.

Hence, proved that cos 60° = 1−tan⁡230∘1+tan⁡230∘=12\dfrac{1 -\tan^2 30^\circ}{1 + \tan^2 30^\circ} = \dfrac{1}{2}.

Question 15

If 0° ≤ x ≤ 90°, state the numerical value of x for which sin x° = cos x°.

Answer

sin 45° = cos 45° = 12\dfrac{1}{\sqrt{2}}.

x = 45 is the only value for which sin x° = cos x°.

Exercise 20(C)

Question 1

Without using trigonometric tables, evaluate :

(i) sin⁡27∘cos⁡63∘\dfrac{\sin 27^\circ}{\cos 63^\circ}

(ii) sec⁡38∘cosec⁡52∘\dfrac{\sec 38^\circ}{\cosec 52^\circ}

(iii) tan⁡19∘cot⁡71∘\dfrac{\tan 19^\circ}{\cot 71^\circ}

Answer

(i) sin⁡27∘cos⁡63∘=sin⁡(90∘−63∘)cos⁡63∘\dfrac{\sin 27^\circ}{\cos 63^\circ} = \dfrac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}

= cos⁡63∘cos⁡63∘\dfrac{\cos 63^\circ}{\cos 63^\circ}

= 1.

(ii) sec⁡38∘cosec⁡52∘\dfrac{\sec 38^\circ}{\cosec 52^\circ}

= sec⁡(90∘−52∘)cosec⁡52∘\dfrac{\sec (90^\circ - 52^\circ)}{\cosec 52^\circ}

= cosec⁡52∘cosec⁡52∘\dfrac{\cosec 52^\circ}{\cosec 52^\circ}

= 1.

(iii) tan⁡19∘cot⁡71∘\dfrac{\tan 19^\circ}{\cot 71^\circ}

= tan⁡(90∘−71∘)cot⁡71∘\dfrac{\tan (90^\circ - 71^\circ)}{\cot 71^\circ}

= cot⁡71∘cot⁡71∘\dfrac{\cot 71^\circ}{\cot 71^\circ}

= 1.

Question 2

Without using trigonometric tables, evaluate :

(i) sin⁡72∘cos⁡18∘−sec⁡32∘cosec⁡58∘\dfrac{\sin 72^\circ}{\cos 18^\circ} - \dfrac{\sec 32^\circ}{\cosec 58^\circ}

(ii) 2tan⁡53∘cot⁡37∘−cot⁡80∘tan⁡10∘\dfrac{2 \tan 53^\circ}{\cot 37^\circ} - \dfrac{\cot 80^\circ}{\tan 10^\circ}

Answer

(i) sin⁡72∘cos⁡18∘−sec⁡32∘cosec⁡58∘\dfrac{\sin 72^\circ}{\cos 18^\circ} - \dfrac{\sec 32^\circ}{\cosec 58^\circ}

= sin⁡(90∘−18∘)cos⁡18∘−sec⁡(90∘−58∘)cosec⁡58∘\dfrac{\sin (90^\circ - 18^\circ)}{\cos 18^\circ} - \dfrac{\sec (90^\circ - 58^\circ)}{\cosec 58^\circ}

= cos⁡18∘cos⁡18∘−cosec⁡58∘cosec⁡58∘\dfrac{\cos 18^\circ}{\cos 18^\circ} - \dfrac{\cosec 58^\circ}{\cosec 58^\circ}

= 1 - 1

= 0.

(ii) 2tan⁡53∘cot⁡37∘−cot⁡80∘tan⁡10∘\dfrac{2 \tan 53^\circ}{\cot 37^\circ} - \dfrac{\cot 80^\circ}{\tan 10^\circ}

= 2tan⁡(90∘−37∘)cot⁡37∘−cot⁡(90∘−10∘)tan⁡10∘\dfrac{2\tan (90^\circ - 37^\circ)}{\cot 37^\circ} - \dfrac{\cot (90^\circ - 10^\circ)}{\tan 10^\circ}

= 2cot⁡37∘cot⁡37∘−tan⁡10∘tan⁡10∘\dfrac{2 \cot 37^\circ}{\cot 37^\circ} - \dfrac{\tan 10^\circ}{\tan 10^\circ}

= 2 - 1

= 1.

Question 3

Without using trigonometric tables, evaluate:

(i) sin⁡80∘cos⁡10∘\dfrac{\sin 80^\circ}{\cos 10^\circ} + sin 59° sec 31°

(ii) cos⁡35∘sin⁡55∘+sin⁡11∘cos⁡79∘\dfrac{\cos 35^\circ}{\sin 55^\circ} + \dfrac{\sin 11^\circ}{\cos 79^\circ} - cos 28° cosec 62°

Answer

(i) sin⁡80∘cos⁡10∘\dfrac{\sin 80^\circ}{\cos 10^\circ} + sin 59° sec 31°

= sin⁡(90∘−10∘)cos⁡10∘\dfrac{\sin (90^\circ - 10^\circ)}{\cos 10^\circ} + sin (90° - 31°) sec 31°

= cos⁡10∘cos⁡10∘\dfrac{\cos10^\circ}{\cos 10^\circ} + cos 31° sec 31°

= 1 + 1sec⁡31∘×sec⁡31∘\dfrac{1}{\sec 31^\circ}\times \sec 31^\circ

= 1 + 1

= 2.

(ii) cos⁡35∘sin⁡55∘+sin⁡11∘cos⁡79∘\dfrac{\cos 35^\circ}{\sin 55^\circ} + \dfrac{\sin 11^\circ}{\cos 79^\circ} - cos 28° cosec 62°

= cos⁡(90∘−55∘)sin⁡55∘+sin⁡(90∘−79∘)cos⁡79∘\dfrac{\cos (90^\circ - 55^\circ)}{\sin 55^\circ} + \dfrac{\sin (90^\circ - 79^\circ)}{\cos 79^\circ} - cos (90°- 62°) cosec 62°

= sin⁡55∘sin⁡55∘+cos⁡79∘cos⁡79∘\dfrac{\sin 55^\circ}{\sin 55^\circ} + \dfrac{\cos 79^\circ}{\cos 79^\circ} - sin 62° cosec 62°

= 1 + 1 - 1cosec⁡62∘×cosec⁡62∘\dfrac{1}{\cosec 62^\circ}\times \cosec 62^\circ

= 1 + 1 - 1

= 1.

Question 4

Without using trigonometric tables, evaluate :

(i) (sin⁡39∘cos⁡51∘)2+(cos⁡51∘sin⁡39∘)2\Big(\dfrac{\sin 39^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos 51^\circ}{\sin 39^\circ}\Big)^2

(ii) (cosec⁡34∘sec⁡56∘)2+(cot⁡81∘tan⁡9∘)2\Big(\dfrac{\cosec 34^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot 81^\circ}{\tan 9^\circ}\Big)^2

Answer

(i) (sin⁡39∘cos⁡51∘)2+(cos⁡51∘sin⁡39∘)2\Big(\dfrac{\sin 39^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos 51^\circ}{\sin 39^\circ}\Big)^2

= (sin⁡(90∘−51∘)cos⁡51∘)2+(cos⁡(90∘−39∘)sin⁡39∘)2\Big(\dfrac{\sin (90^\circ - 51^\circ)}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\cos (90^\circ - 39^\circ)}{\sin 39^\circ}\Big)^2

= (cos⁡51∘cos⁡51∘)2+(sin⁡39∘sin⁡39∘)2\Big(\dfrac{\cos 51^\circ}{\cos 51^\circ}\Big)^2 + \Big(\dfrac{\sin 39^\circ}{\sin 39^\circ}\Big)^2

= 12 + 12

= 2.

(ii) (cosec⁡34∘sec⁡56∘)2+(cot⁡81∘tan⁡9∘)2\Big(\dfrac{\cosec 34^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot 81^\circ}{\tan 9^\circ}\Big)^2

= (cosec⁡(90∘−56∘)sec⁡56∘)2+(cot⁡(90∘−9∘)tan⁡9∘)2\Big(\dfrac{\cosec (90^\circ -56^\circ)}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\cot (90^\circ - 9^\circ)}{\tan 9^\circ}\Big)^2

= (sec⁡56∘sec⁡56∘)2+(tan⁡9∘tan⁡9∘)2\Big(\dfrac{\sec 56^\circ}{\sec 56^\circ}\Big)^2 + \Big(\dfrac{\tan 9^\circ}{\tan 9^\circ}\Big)^2

= 1 + 1

= 2.

Question 5

Without using trigonometric tables, evaluate:

(i) 3 cos 80° cosec 10° + 2 cos 59°cosec 31°

(ii) 5 sin 70° sec 20° - 3 sin 50° sec 40°

Answer

(i) 3 cos 80° cosec 10° + 2 cos 59° cosec 31°

= 3 cos (90° - 10°) cosec 10° + 2 cos (90° - 31°) cosec 31°

= 3 sin 10° cosec 10° + 2 sin 31°cosec 31°

= 3×1cosec⁡10∘×cosec⁡10∘+2×1cosec⁡31∘×cosec⁡31∘3\times\dfrac{1}{\cosec 10^\circ}\times \cosec 10^\circ + 2\times \dfrac{1}{\cosec 31^\circ}\times \cosec 31^\circ

= 3 + 2

= 5.

(ii) 5 sin 70° sec 20° - 3 sin 50° sec 40°

= 5 sin (90° - 20°) sec 20° - 3 sin (90° - 40°) sec 40°

= 5 cos 20° sec 20° - 3 cos 40° sec 40°

= 5×1sec⁡20∘×sec⁡20∘−3×1sec⁡40∘×sec⁡40∘5\times\dfrac{1}{\sec 20^\circ}\times \sec 20^\circ - 3\times \dfrac{1}{\sec 40^\circ}\times \sec 40^\circ

= 5 - 3

= 2.

Question 6

Without using trigonometric tables, evaluate:

(i) sin⁡35∘cos⁡55∘+cos⁡35∘sin⁡55∘cosec⁡210∘−tan⁡280∘\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ}

(ii) 4 tan 60° sec 30° + sin⁡31∘sec⁡59∘+cot⁡59∘cot⁡31∘8sin⁡230∘−tan⁡245∘\dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}

Answer

(i) Solving,

sin⁡35∘cos⁡55∘+cos⁡35∘sin⁡55∘cosec⁡210∘−tan⁡280∘=sin⁡(90∘−55∘)cos⁡55∘+cos⁡(90∘−55∘)sin⁡55∘cosec⁡2(90∘−80∘)−tan⁡280∘=cos⁡55∘cos⁡55∘+sin⁡55∘sin⁡55∘sec⁡280∘−tan⁡280∘=cos⁡255∘+sin⁡255∘1=1.\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ}\\[1em] = \dfrac{\sin (90^\circ - 55^\circ) \cos 55^\circ + \cos (90^\circ - 55^\circ) \sin 55^\circ}{\cosec^2 (90^\circ - 80^\circ) - \tan^2 80^\circ}\\[1em] = \dfrac{\cos 55^\circ \cos 55^\circ + \sin 55^\circ \sin 55^\circ}{\sec^2 80^\circ - \tan^2 80^\circ}\\[1em] = \dfrac{\cos^2 55^\circ + \sin^2 55^\circ}{1} \\[1em] = 1.

Hence, sin⁡35∘cos⁡55∘+cos⁡35∘sin⁡55∘cosec⁡210∘−tan⁡280∘=1.\dfrac{\sin 35^\circ \cos 55^\circ + \cos 35^\circ \sin 55^\circ}{\cosec^2 10^\circ - \tan^2 80^\circ} = 1.

(ii) Solving,

⇒4×tan⁡60∘sec⁡30∘+sin⁡31∘sec⁡59∘+cot⁡59∘cot⁡31∘8sin⁡230∘−tan⁡245∘⇒4×sin⁡60∘cos⁡60∘×1cos⁡30∘+sin⁡(90∘−59∘)1cos⁡59∘+cot⁡(90∘−31∘)cot⁡31∘8sin⁡230∘−sin⁡245∘cos⁡245∘⇒4×sin⁡(90∘−30∘)cos⁡60∘×1cos⁡30∘+cos⁡59∘×1cos⁡59∘+tan⁡31∘cot⁡31∘8sin⁡230∘−sin⁡2(90∘−45∘)cos⁡245∘⇒4×cos⁡30∘cos⁡60∘×1cos30∘+1+18sin⁡230∘−cos⁡245∘cos⁡245∘⇒4×1cos⁡60∘+1+18sin⁡230∘−1⇒4×sec⁡60∘+28×14−1⇒4×2+22−1⇒8+2⇒10.\Rightarrow 4 \times \tan 60^\circ \sec 30^\circ + \dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ}\\[1em] \Rightarrow 4 \times\dfrac{\sin 60^\circ}{\cos 60^\circ} \times \dfrac{1}{\cos 30^\circ} + \dfrac{\sin (90^\circ - 59^\circ) \dfrac{1}{\cos 59^\circ }+ \cot (90^\circ - 31^\circ) \cot 31^\circ}{8 \sin^2 30^\circ - \dfrac{\sin^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{\sin (90^\circ -30^\circ)}{\cos 60^\circ} \times \dfrac{1}{\cos 30^\circ} + \dfrac{\cos 59^\circ \times \dfrac{1}{\cos 59^\circ }+ \tan 31^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \dfrac{\sin^2 (90^\circ - 45^\circ)}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{\cos 30^\circ}{\cos 60^\circ} \times \dfrac{1}{cos 30^\circ} + \dfrac{1 + 1}{8 \sin^2 30^\circ - \dfrac{\cos^2 45^\circ}{\cos^2 45^\circ}}\\[1em] \Rightarrow 4 \times\dfrac{1}{\cos 60^\circ} + \dfrac{1 + 1}{8 \sin^2 30^\circ - 1}\\[1em] \Rightarrow 4 \times \sec 60^\circ + \dfrac{2}{8 \times \dfrac{1}{4} - 1}\\[1em] \Rightarrow 4 \times 2 + \dfrac{2}{2 - 1}\\[1em] \Rightarrow 8 + 2 \\[1em] \Rightarrow 10.

Hence, 4 tan 60° sec 30° + sin⁡31∘sec⁡59∘+cot⁡59∘cot⁡31∘8sin⁡230∘−tan⁡245∘\dfrac{\sin 31^\circ \sec 59^\circ + \cot 59^\circ \cot 31^\circ}{8 \sin^2 30^\circ - \tan^2 45^\circ} = 10.

Question 7

Without using trigonometric tables, evaluate:

(i) sin⁡29∘31′cos⁡60∘29′\dfrac{\sin 29^\circ 31'}{\cos 60^\circ 29'}

(ii) tan⁡20∘43′cot⁡69∘17′\dfrac{\tan 20^\circ 43'}{\cot 69^\circ 17'}

(iii) sin⁡28∘54′sec⁡61∘6′+cos⁡28∘54′cosec⁡61∘6′\dfrac{\sin 28^\circ 54'}{\sec 61^\circ 6'} + \dfrac{\cos 28^\circ 54'}{\cosec 61^\circ 6'}

Answer

(i) sin⁡29∘31′cos⁡60∘29′\dfrac{\sin 29^\circ 31'}{\cos 60^\circ 29'}

= sin⁡(90∘−60∘29′)cos⁡60∘29′\dfrac{\sin (90^\circ - 60^\circ 29')}{\cos 60^\circ 29'}

= cos⁡60∘29′cos⁡60∘29′\dfrac{\cos 60^\circ 29'}{\cos 60^\circ 29'}

= 1.

(ii) tan⁡20∘43′cot⁡69∘17′\dfrac{\tan 20^\circ 43'}{\cot 69^\circ 17'}

= tan⁡(90∘−69∘17′)cot⁡69∘17′\dfrac{\tan (90^\circ - 69^\circ 17')}{\cot 69^\circ 17'}

= cot⁡69∘17′cot⁡69∘17′\dfrac{\cot 69^\circ 17'}{\cot 69^\circ 17'}

= 1.

(iii) sin⁡28∘54′sec⁡61∘6′+cos⁡28∘54′cosec⁡61∘6′\dfrac{\sin 28^\circ 54'}{\sec 61^\circ 6'} + \dfrac{\cos 28^\circ 54'}{\cosec 61^\circ 6'}

= sin⁡(90∘−61∘6′)sec⁡61∘6′+cos⁡(90∘−61∘6′)cosec⁡61∘6′\dfrac{\sin (90^\circ - 61^\circ 6')}{\sec 61^\circ 6'} + \dfrac{\cos (90^\circ - 61^\circ 6')}{\cosec 61^\circ 6'}

= cos⁡61∘6′sec⁡61∘6′+sin⁡61∘6′cosec⁡61∘6′\dfrac{\cos 61^\circ 6'}{\sec 61^\circ 6'} + \dfrac{\sin 61^\circ 6'}{\cosec 61^\circ 6'}

= cos 61°6' cos 61°6' + sin 61°6' sin 61°6'

= cos2 61°6' + sin2 61°6'

= 1.

Question 8

Without using trigonometric tables, evaluate:

(i) sin 37° - cos 53°

(ii) cos2 25° - sin2 65°

(iii) sin2 23° + sin2 67°

Answer

(i) sin 37° - cos 53°

= sin(90° - 53°) - cos 53°

= cos 53° - cos 53°

= 0.

(ii) cos2 25° - sin2 65°

= cos2 (90° -65°) - sin2 65°

= sin2 65° - sin2 65°

= 0.

(iii) sin2 23° + sin2 67°

= sin2 (90° - 67°) + sin2 67°

= cos2 67° + sin2 67°

= 1.

Question 9

Without using trigonometric tables, evaluate :

(i) sec2 36° - cot2 54°

(ii) cosec2 38° - tan2 52°

(iii) cos2 24° + cos2 66°

Answer

(i) sec2 36° - cot2 54°

= sec2 (90° - 54°) - cot2 54°

= cosec2 54° - cot2 54°

= 1.

(ii) cosec2 38° - tan2 52°

= cosec2 (90° - 52°) - tan2 52°

= sec2 52° - tan2 52°

= 1.

(iii) cos2 24° + cos2 66°

= cos2 (90° - 66°) + cos2 66°

= sin2 66° + cos2 66°

= 1.

Question 10

Without using trigonometric tables, evaluate:

(i) cos⁡234∘+cos⁡256∘sin⁡259∘+sin⁡231∘\dfrac{\cos^2 34^\circ + \cos^2 56^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}

(ii) cosec⁡267∘−tan⁡223∘sec⁡220∘−cot⁡270∘\dfrac{\cosec^2 67^\circ - \tan^2 23^\circ}{\sec^2 20^\circ - \cot^2 70^\circ}

Answer

(i) cos⁡234∘+cos⁡256∘sin⁡259∘+sin⁡231∘\dfrac{\cos^2 34^\circ + \cos^2 56^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}

= cos⁡2(90∘−56∘)+cos⁡256∘sin⁡2(90∘−31∘)+sin⁡231∘\dfrac{\cos^2 (90^\circ - 56^\circ) + \cos^2 56^\circ}{\sin^2 (90^\circ - 31^\circ) + \sin^2 31^\circ}

= sin⁡256∘+cos⁡256∘cos⁡231∘+sin⁡231∘\dfrac{\sin^2 56^\circ + \cos^2 56^\circ}{\cos^2 31^\circ + \sin^2 31^\circ}

= 1.

(ii) cosec⁡267∘−tan⁡223∘sec⁡220∘−cot⁡270∘\dfrac{\cosec^2 67^\circ - \tan^2 23^\circ}{\sec^2 20^\circ - \cot^2 70^\circ}

= cosec⁡2(90∘−23∘)−tan⁡223∘sec⁡2(90∘−70∘)−cot⁡270∘\dfrac{\cosec^2 (90^\circ - 23^\circ) - \tan^2 23^\circ}{\sec^2 (90^\circ - 70^\circ) - \cot^2 70^\circ}

= sec⁡223∘−tan⁡223∘cosec⁡270∘−cot⁡270∘\dfrac{\sec^2 23^\circ - \tan^2 23^\circ}{\cosec^2 70^\circ - \cot^2 70^\circ}

= 1.

Question 11

Without using trigonometric tables, prove that :

(i) sin 73° cos 17° + cos 73° sin 17° = 1

(ii) sin 40° sec 50° + cos 40° cosec 50° = 2

(iii) sec275° - cot215° = 1

(iv) cos223° + cos267° = 1

(v) cosec256° - tan234° = 1

(vi) tan 10° tan 15° tan 75° tan 80° = 1

Answer

(i) sin 73° cos 17° + cos 73° sin 17° = 1

Solving L.H.S.,

sin 73° cos 17° + cos 73° sin 17°

= sin (90° - 17°) cos 17° + cos (90° - 17°) sin 17°

= cos 17° cos 17° + sin 17° sin 17°

= cos2 17° + sin2 17°

= 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 73° cos 17° + cos 73° sin 17° = 1.

(ii) sin 40° sec 50° + cos 40° cosec 50° = 2

Solving L.H.S.

sin (90° - 50°) sec 50° + cos (90° - 50°) cosec 50°

= cos 50° sec 50° + sin 50° cosec 50°

= 1sec⁡50∘×sec⁡50∘+1cosec⁡50∘×cosec⁡50∘\dfrac{1}{\sec 50^\circ}\times \sec 50^\circ + \dfrac{1}{\cosec 50^\circ}\times \cosec 50^\circ

= 1 + 1 = 2

Since, L.H.S. = R.H.S.

Hence, proved that sin 40° sec 50° + cos 40° cosec 50° = 2.

(iii) sec275° - cot215° = 1

Solving L.H.S.

= sec275° - cot215°

= sec2(90° - 15°) - cot215°

= cosec215° - cot215°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that sec275° - cot215° = 1.

(iv) cos223° + cos267° = 1

Solving L.H.S.

cos223° + cos267°

= cos2(90° - 67°) + cos267°

= sin267° + cos267°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that cos223° + cos267° = 1.

(v) cosec256° - tan234° = 1

Solving L.H.S.

cosec256° - tan234°

= cosec2(90° - 34°) - tan234°

= sec234° - tan234°

= 1

Since, L.H.S. = R.H.S.

Hence, proved that cosec256° - tan234° = 1.

(vi) tan 10° tan 15° tan 75° tan 80° = 1

Solving L.H.S.

tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90°- 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= 1tan⁡80∘×tan⁡80∘×1tan⁡75∘×tan⁡75∘\dfrac{1}{\tan80^\circ}\times \tan 80^\circ \times \dfrac{1}{\tan 75^\circ}\times \tan 75^\circ

= 1

Since, L.H.S. = R.H.S.

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

Question 12

Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.

(i) cos 81° + cot 81°

(ii) cos 76° + sec 76°

(iii) sec 63° + cosec 49°

(iv) sin 59° + cos 56°

Answer

(i) cos 81° + cot 81°

= cos (90° - 9°) + cot (90° - 9°)

= sin 9° + tan 9°

Hence, cos 81° + cot 81° = sin 9° + tan 9°.

(ii) cos 76° + sec 76°

= cos (90° - 14°) + sec (90° - 14°)

= sin 14° + cosec 14°

Hence, cos 76° + sec 76° = sin 14° + cosec 14°.

(iii) sec 63° + cosec 49°

= sec (90° - 27°) + cosec (90° - 41°)

= cosec 27° + sec 41°

Hence, sec 63° + cosec 49° = cosec 27° + sec 41°.

(iv) sin 59° + cos 56°

= sin (90° - 31°) + cos (90° - 34°)

= cos 31° + sin 34°

Hence, sin 59° + cos 56° = cos 31° + sin 34°.

Question 13

If 0° < θ < 25°, prove that cos(65° + θ) - sin (25° - θ) = 0.

Answer

As,

cos θ = sin (90° - θ)

Solving,

cos(65° + θ) - sin (25° - θ)

= sin (90° - (65° + θ)) - sin(25° - θ)

= sin (90° - 65° - θ) - sin(25° - θ)

= sin (25° - θ) - sin (25° - θ)

= 0.

Hence, proved that cos(65° + θ) - sin (25° - θ) = 0.

Question 14

Prove that :

sin (50° + θ) - cos (40° - θ) = 0

Answer

Solving,

sin (50° + θ) - cos (40° - θ)

= cos (90° - (50° + θ)) - cos (40° - θ)

= cos (90° - 50° - θ) - cos (40° - θ)

= cos (40° - θ) - cos (40° - θ)

= 0.

Hence, proved that sin (50° + θ) - cos (40° - θ) = 0.

Question 15

Prove that : tan(45° - A) tan(45° + A) = 1.

Answer

Solving,

tan(45° - A) tan(45° + A)

= tan[90° - (45° + A)] tan(45° + A)

= cot(45° + A) tan(45° + A)

= 1tan⁡(45∘+A)×tan⁡(45∘+A)\dfrac{1}{\tan(45^\circ + A)}\times \tan(45^\circ + A)

= 1.

Hence, proved that tan(45° - A) tan(45° + A) = 1.

Question 16

Prove that :

(i) cos⁡Asin⁡(90∘−A)+sin⁡Acos⁡(90∘−A)\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)} = 2

(ii) sin⁡Asin⁡(90∘−A)+cos⁡Acos⁡(90∘−A)\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)} = sec(90° - A) cosec(90° - A)

(iii) sin(90° - A) cos(90° - A) = tan⁡A1+tan⁡2A\dfrac{\tan A}{1 + \tan^2 A}

Answer

(i) cos⁡Asin⁡(90∘−A)+sin⁡Acos⁡(90∘−A)\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)} = 2

Solving L.H.S.,

cos⁡Asin⁡(90∘−A)+sin⁡Acos⁡(90∘−A)\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)}

= cos⁡Acos⁡A+sin⁡Asin⁡A\dfrac{\cos A}{\cos A} + \dfrac{\sin A}{\sin A}

= 1 + 1

= 2.

Hence, proved that cos⁡Asin⁡(90∘−A)+sin⁡Acos⁡(90∘−A)\dfrac{\cos A}{\sin(90^\circ - A)} + \dfrac{\sin A}{\cos (90^\circ - A)} = 2.

(ii) sin⁡Asin⁡(90∘−A)+cos⁡Acos⁡(90∘−A)\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)} = sec(90° - A) cosec(90° - A)

Solving L.H.S.,

⇒sin⁡Asin⁡(90∘−A)+cos⁡Acos⁡(90∘−A)=sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡2A+cos⁡2Acos⁡Asin⁡A=1sin A cos A=cosec⁡Asec⁡A=sec⁡(90∘−A)cosec⁡(90∘−A).\Rightarrow\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)}\\[1em] = \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A} \\[1em] = \dfrac{\sin^2 A + \cos^2 A}{\cos A \sin A} \\[1em] = \dfrac{1}{\text{sin A cos A}}\\[1em] = \cosec A \sec A\\[1em] = \sec(90^\circ - A) \cosec(90^\circ - A).

Hence, proved that sin⁡Asin⁡(90∘−A)+cos⁡Acos⁡(90∘−A)\dfrac{\sin A}{\sin(90^\circ - A)} + \dfrac{\cos A}{\cos (90^\circ - A)} = sec(90° - A) cosec(90° - A).

(iii) sin(90° - A) cos(90° - A) = tan⁡A1+tan⁡2A\dfrac{\tan A}{1 + \tan^2 A}

Solving L.H.S.,

sin(90° - A) cos(90° - A)

= cos A sin A

Solving R.H.S.,

⇒tan⁡A1+tan⁡2A=sin⁡Acos⁡A1+sin⁡2Acos⁡2A=sin⁡Acos⁡Acos⁡2A+sin⁡2Acos⁡2A=sin⁡Acos⁡A1cos⁡2A=sin⁡Acos⁡A×cos⁡2A=sin⁡Acos⁡A.\Rightarrow \dfrac{\tan A}{1 + \tan^2 A} \\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{1+ \dfrac{\sin^2 A}{\cos^2 A}}\\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{ \dfrac{\cos^2 A + \sin^2 A}{\cos^2 A}}\\[1em] = \dfrac{\dfrac{\sin A}{\cos A}}{ \dfrac{1}{\cos^2 A}}\\[1em] = \dfrac{\sin A}{\cos A}\times\cos^2 A\\[1em] = \sin A \cos A.

Since, L.H.S. = R.H.S. = sin A. cos A

Hence, proved that sin(90° - A) cos(90° - A) = tan⁡A1+tan⁡2A\dfrac{\tan A}{1 + \tan^2 A}.

Exercise 20(D)

Question 1

A balloon is connected to a meterological station by a cable of length 200 metres, inclined at 60° to the horizontal. Determine the height of the balloon from the ground, assuming that there is no slack in the string. (Take 3\sqrt{3} = 1.73)

Answer

A balloon is connected to a meterological station by a cable of length 200 metres, inclined at 60 to the horizontal. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Let A be the position of balloon and C be the position of meterological station. Then length of cable (AC) = 200 m.

Let height of the ballon from the ground = x

In triangle ABC,

sin 60° = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

32=x200\dfrac{\sqrt{3}}{2} = \dfrac{x}{200}

x = 20032\dfrac{200 \sqrt{3}}{2}

x = 1003100\sqrt{3}

x = 100(1.73)

x = 173 m.

Hence, height of the balloon from the ground = 173 m.

Question 2

A ladder leaning against a wall, makes an angle of 60° with the horizontal and the foot of tha ladder is 9.5 metres away from the wall. Find the length of the ladder.

Answer

Let AB be the wall and AC be the ladder.

A ladder leaning against a wall, makes an angle of 60 with the horizontal and the foot of tha ladder is 9.5 metres away from the wall. Find the length of the ladder. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

In triangle ABC,

⇒cos⁡60∘=BCAC⇒12=9.5AC⇒AC=2×9.5=19 m.\Rightarrow \cos 60^\circ = \dfrac{BC}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{9.5}{AC} \\[1em] \Rightarrow AC = 2 \times 9.5 = 19 \text{ m}.

Hence, length of ladder = 19 m.

Question 3

A kite is flying with a thread 150 m long. If the thread is assumed stretched straight and makes an angle of 60° with the horizontal, find the height of the kite above the ground. (Take 3\sqrt{3} = 1.73)

Answer

A kite is flying with a thread 150 m long. If the thread is assumed stretched straight and makes an angle of 60. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Let position of kite be A and C be the point on ground.

AC = length of string = 150 m.

AB = Height of the kite above the ground

sin 60° = ABAC\dfrac{AB}{AC}

32=AB150\dfrac{\sqrt{3}}{2} = \dfrac{AB}{150}

AB = 753=75×1.7375\sqrt{3} = 75 \times 1.73 = 129.75 m

Hence, height of the kite above ground = 129.75 m.

Question 4

A kite is flying at a height of 120 m from the level ground. It is attached to a string inclined at 60° to the horizontal. Find the length of the string. (Take 3\sqrt{3} = 1.73)

Answer

A kite is flying at a height of 120 m from the level ground. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Let A be the position of kite and C be the point from where string is attached.

AB = 120 m

In triangle ABC,

⇒sin⁡60∘=ABAC⇒32=120AC⇒AC=120×23⇒AC=2403⇒AC=240×33×3⇒AC=24033⇒AC=803=138.4 m.\Rightarrow \sin 60^\circ = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{120}{AC} \\[1em] \Rightarrow AC = \dfrac{120 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240\sqrt{3}}{3} \\[1em] \Rightarrow AC = 80\sqrt{3} = 138.4 \text{ m}.

Hence, length of the string = 138.4 m.

Question 5

In a △ABC, right angled at B, if ∠A = 30° and BC = 8 cm, find the remaining angles and sides.

Answer

In a △ABC, right angled at B, if ∠A = 30&deg; and BC = 8 cm, find the remaining angles and sides. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Given,

∠A = 30°, BC = 8 cm

sin 30° = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

12=BCAC\dfrac{1}{2} = \dfrac{BC}{AC}

12=8AC\dfrac{1}{2} = \dfrac{8}{AC}

AC = 16 cm

Now,

cos 30° = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

32=ABAC\dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC}

32=AB16\dfrac{\sqrt{3}}{2} = \dfrac{AB}{16}

AB = 838\sqrt{3} cm

By angle sum property of triangle,

∠A + ∠B + ∠C = 180°

30° + 90° + ∠C = 180°

∠C + 120° = 180°

∠C = 180° - 120° = 60°

Hence, ∠C = 60°, AC = 16 cm and AB = 838\sqrt{3} cm.

Question 6

In a rectangle ABCD, AB = 12 cm and ∠BAC = 30°. Calculate the lengths of side BC and diagonal AC.

Answer

In a rectangle ABCD, AB = 12 cm and ∠BAC = 30. Calculate the lengths of side BC and diagonal AC. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

In △BAC,

Base = AB = 12 cm and ∠BAC = 30°

Perpendicular = BC

tan 30° = BCAB\dfrac{BC}{AB}

13=BC12\dfrac{1}{\sqrt{3}} = \dfrac{BC}{12}

BC = 123=43\dfrac{12}{\sqrt{3}} = 4\sqrt{3} cm

Now,

cos 30° = ABAC\dfrac{AB}{AC}

32=12AC\dfrac{\sqrt{3}}{2} = \dfrac{12}{AC}

AC = 243=83\dfrac{24}{\sqrt{3}} = 8\sqrt{3} cm.

Hence, BC = 434\sqrt{3} cm and AC = 838\sqrt{3} cm.

Multiple Choice Questions

Question 1

If θ is an acute angle and sin(θ - 15°) = 12\dfrac{1}{2}, then cos(θ - 15°) =

  1. 12\dfrac{1}{2}

  2. 32\dfrac{\sqrt{3}}{2}

  3. 12\dfrac{1}{\sqrt{2}}

  4. 1

Answer

Given,

sin(θ - 15°)= 12\dfrac{1}{2}

sin(θ - 15°) = sin 30°

θ - 15° = 30°

θ = 30° + 15° = 45°

so,

cos(θ - 15°) = cos(45° - 15°) = cos 30° = 32\dfrac{\sqrt{3}}{2}

Hence, option 2 is the correct option.

Question 2

If 0° ≤θ ≤ 90° and cos(θ - 30°) = 12\dfrac{1}{2}, then tan θ =

  1. 3\sqrt{3}

  2. 1

  3. 12\dfrac{1}{\sqrt{2}}

  4. undefined

Answer

Given,

cos(θ - 30°) = 12\dfrac{1}{2}

cos(θ - 30°) = cos 60°

θ - 30° = 60°

θ = 60° + 30° = 90°

Then,

tan θ = tan 90° = undefined.

Hence, option 4 is the correct option.

Question 3

If x tan 30° = cos 60°, then x =

  1. 2

  2. 23\dfrac{2}{\sqrt{3}}

  3. 32\dfrac{\sqrt{3}}{2}

  4. 12\dfrac{1}{2}

Answer

Given,

x tan 30° = cos 60°

x×13=12x \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{2}

x = 32\dfrac{\sqrt{3}}{2}

Hence, option 3 is the correct option.

Question 4

If 0° ≤θ ≤ 90° and tan (θ + 15°)= 1, then cos 2θ =

  1. 12\dfrac{1}{2}

  2. 32\dfrac{\sqrt{3}}{2}

  3. 12\dfrac{1}{\sqrt{2}}

  4. 0

Answer

tan (θ + 15°) = 1

tan (θ + 15°) = tan 45°

θ + 15° = 45°

θ = 45° - 15°= 30°

Then

cos 2θ = cos 2(30°) = cos 60° = 12\dfrac{1}{2}.

Hence, option 1 is the correct option.

Question 5

If sin θ = cos θ, then sec (θ + 15°) =

  1. 2\sqrt{2}

  2. 2

  3. 23\dfrac{2}{\sqrt{3}}

  4. 1

Answer

Given,

sin θ = cos θ

This is possible in case of θ = 45° as sin 45° = cos 45° = 12\dfrac{1}{\sqrt{2}}.

θ = 45°

Then,

sec (θ + 15°) = sec (45° + 15°)

= sec 60°

= 2.

Hence, option 2 is the correct option.

Question 6

If cos 2θ = 0 and θ is an acute angle, then cot(θ - 15°) =

  1. 13\dfrac{1}{\sqrt{3}}

  2. 1

  3. 3\sqrt{3}

  4. undefined

Answer

Given,

cos 2θ = 0

cos 2θ = cos 90°

2θ = 90°

θ = 45°

Then,

cot(θ - 15°) = cot (45° - 15°) = cot 30° = 3\sqrt{3}.

Hence, option 3 is the correct option.

Question 7

If θ is an acute angle and sin (θ + 18°) = 12\dfrac{1}{2}, then cosec 5θ =

  1. 2

  2. 2\sqrt{2}

  3. 1

  4. 23\dfrac{2}{\sqrt{3}}

Answer

Given,

sin(θ + 18°) = 12\dfrac{1}{2}

sin(θ + 18°) = sin 30°

θ + 18° = 30°

θ = 30° - 18° = 12°

Then,

cosec 5θ = cosec 5(12°) = cosec 60° = 23\dfrac{2}{\sqrt{3}}.

Hence, option 4 is the correct option.

Question 8

If sin θ = 817\dfrac{8}{17}, then cot θ =

  1. 158\dfrac{15}{8}

  2. 1517\dfrac{15}{17}

  3. 815\dfrac{8}{15}

  4. 178\dfrac{17}{8}

Answer

sin θ = 817\dfrac{8}{17}

sin θ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

Let Perpendicular = 8x and Hypotenuse = 17x

We will find Base by using Pythagoras Theorem,

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

Base2 = (17x)2 - (8x)2

Base2 = 289x2 - 64x2

Base2 = 225x2

Base = 15x

Then,

cot θ = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= 15x8x=158\dfrac{15x}{8x} = \dfrac{15}{8}.

Hence, option 1 is the correct option.

Question 9

If sin θ = 12\dfrac{1}{2}, then (3cos θ - 4 cos3 θ)=

  1. 0

  2. 12\dfrac{1}{2}

  3. 16\dfrac{1}{6}

  4. -1

Answer

sin θ = 12\dfrac{1}{2}

sin 30° = 12\dfrac{1}{2}

sin θ = sin 30°

θ = 30°

Then,

3cos θ - 4 cos3 θ = 3 cos 30° - 4 (cos 30°)3

= 3×32−4×(32)33\times \dfrac{\sqrt{3}}{2}- 4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3

= 332−1238\dfrac{3\sqrt{3}}{2}- \dfrac{12\sqrt{3}}{8}

= 332−332\dfrac{3\sqrt{3}}{2}- \dfrac{3\sqrt{3}}{2}

= 0.

Hence, option 1 is the correct option.

Question 10

If 5 cot θ = 3, then (5sin⁡θ−3cos⁡θ)(4sin⁡θ+3cos⁡θ)\dfrac{(5 \sin θ - 3 \cos θ)}{(4 \sin θ + 3 \cos θ)} =

  1. 1118\dfrac{11}{18}

  2. 1629\dfrac{16}{29}

  3. 1427\dfrac{14}{27}

  4. 2916\dfrac{29}{16}

Answer

Given,

5 cot θ = 3

cot θ = 35\dfrac{3}{5}

cot θ = cos⁡θsin⁡θ\dfrac{\cos θ}{\sin θ}

Given,

(5sin⁡θ−3cos⁡θ)(4sin⁡θ+3cos⁡θ)\dfrac{(5 \sin θ - 3 \cos θ)}{(4 \sin θ + 3 \cos θ)}

Dividing above equation by sin θ, we get :

⇒5sin⁡θ−3cos⁡θsin⁡θ4sin⁡θ+3cos⁡θsin⁡θ⇒(5−3cot⁡θ)(4+3cot⁡θ)=(5−3×35)(4+3×35)=(5−95)(4+95)=(25−95)(20+95)=(165)(295)=165×529=1629.\Rightarrow \dfrac{\dfrac{5\sin θ - 3\cos θ}{\sin θ}}{\dfrac{4\sin θ + 3 \cos θ}{\sin θ}} \\[1em] \Rightarrow \dfrac{(5 - 3 \cot θ)}{(4 + 3 \cot θ)}\\[1em] = \dfrac{\Big(5 - 3\times \dfrac{3}{5}\Big)}{\Big(4 + 3\times \dfrac{3}{5}\Big)}\\[1em] = \dfrac{\Big(5 - \dfrac{9}{5}\Big)}{\Big(4 + \dfrac{9}{5}\Big)}\\[1em] = \dfrac{\Big(\dfrac{{25 -9}}{5}\Big)}{\Big( \dfrac{20 + 9}{5}\Big)}\\[1em] = \dfrac{\Big(\dfrac{{16}}{5}\Big)}{\Big( \dfrac{29}{5}\Big)}\\[1em] = \dfrac{16}{5}\times \dfrac{5}{29}\\[1em] = \dfrac{16}{29}.

Hence, option 2 is the correct option.

Question 11

In △ABC, ∠B = 90°, AB = 5 cm and BC = 12 cm. Then sin C =

  1. 1213\dfrac{12}{13}

  2. 513\dfrac{5}{13}

  3. 512\dfrac{5}{12}

  4. 135\dfrac{13}{5}

AB = 5 cm and BC = 12 cm. Then sin C. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Answer

Perpendicular = AB = 5 cm

Base = BC = 12 cm

By using Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

Hypotenuse2 = (5)2 + (12)2

Hypotenuse2 = 252 + 1442

Hypotenuse2 = 1692

Hypotenuse = 13 cm

AC = 13 cm

sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

= 513\dfrac{5}{13}.

Hence, option 2 is the correct option.

Question 12

The value of sin θ cos (90° - θ) + cos θ sin(90° - θ) =

  1. 0

  2. 1

  3. 2

  4. 32\dfrac{3}{2}

Answer

Solving,

⇒ sin θ cos (90° - θ) + cos θ sin(90° - θ)

⇒ sin θ sin θ + cos θ cos θ

⇒ sin2θ + cos2θ

⇒ 1.

Hence, option 2 is the correct option.

Question 13

The value of sin225° + sin265° =

  1. 90

  2. 40

  3. 0

  4. 1

Answer

⇒ sin225° + sin265°

⇒ sin225° + sin2(90° - 25°)

⇒ sin225° + cos225°

⇒ 1.

Hence, option 4 is the correct option.

Question 14

The value of tan⁡35∘cot⁡55∘+cot⁡78∘tan⁡12∘\dfrac{\tan 35^\circ}{\cot 55^\circ} + \dfrac{\cot 78^\circ}{\tan 12^\circ} =

  1. 0

  2. 2

  3. 1

  4. 3

Answer

Given,

⇒tan⁡35∘cot⁡55∘+cot⁡78∘tan⁡12∘\Rightarrow \dfrac{\tan 35^\circ}{\cot 55^\circ} + \dfrac{\cot 78^\circ}{\tan 12^\circ}

= tan⁡35∘cot⁡(90∘−35∘)+cot⁡(90∘−12∘)tan⁡12∘\dfrac{\tan 35^\circ}{\cot (90^\circ - 35^\circ)} + \dfrac{\cot (90^\circ - 12^\circ)}{\tan 12^\circ}

= tan⁡35∘tan⁡35∘+tan⁡12∘tan⁡12∘\dfrac{\tan 35^\circ}{\tan 35^\circ} + \dfrac{\tan 12^\circ}{\tan 12^\circ}

= 1 + 1 = 2.

Hence, option 2 is the correct option.

Question 15

If 3 sin θ + 4 cos θ = 5, then the value of sin θ is :

  1. 34\dfrac{3}{4}

  2. 35\dfrac{3}{5}

  3. 45\dfrac{4}{5}

  4. 53\dfrac{5}{3}

Answer

3 sin θ + 4 cos θ = 5

4 cos θ = 5 - 3 sin θ

Squaring Both Sides,

(4 cos θ)2 = (5 - 3 sin θ)2

16 cos2θ = 25 + 9 sin2θ - 30 sin θ

Putting cos2θ = 1 - sin2θ

16 ( 1 - sin2θ) = 25 + 9 sin2θ - 30 sin θ

16 - 16 sin2θ = 25 + 9 sin2θ - 30 sin θ

25 + 9 sin2θ - 30 sin θ - 16 + 16 sin2θ = 0

25 sin2θ - 30 sin θ + 9 = 0

25 sin2θ - 15 sin θ - 15 sin θ + 9 = 0

5 sin θ(5sin θ - 3) - 3(5sin θ - 3) = 0

(5 sin θ - 3)(5sin θ - 3)= 0

(5 sin θ - 3)2 = 0

5 sin θ - 3 = 0

⇒sin⁡θ=35\Rightarrow \sin θ = \dfrac{3}{5}

Hence, option 2 is the correct option.

Question 16

The value of tan 5° tan 25° tan 30° tan 65° tan 85° =

  1. 1

  2. 3\sqrt{3}

  3. 13\dfrac{1}{\sqrt{3}}

  4. 2

Answer

Solving,

⇒ tan 5° tan 25° tan 30° tan 65° tan 85°

⇒ tan 5° tan 85° tan 25° tan 65° tan 30°

⇒ tan 5° tan (90° - 5°) tan 25° tan (90° - 25°) tan 30°

⇒ tan 5° cot 5° tan 25° cot 25° tan 30°

⇒ 1×1×1 \times 1 \times tan 30°

⇒ 13\dfrac{1}{\sqrt{3}}.

Hence, option 3 is the correct option.

Question 17

The value of (cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°) =

  1. 35\dfrac{3}{5}

  2. 56\dfrac{5}{6}

  3. 74\dfrac{7}{4}

  4. 58\dfrac{5}{8}

Answer

(cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)

⇒(1+12+12)(1−12+12)=(32+12)(32−12)=(32)2−(12)2=94−12=9−24=74\Rightarrow \Big(1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big) \Big(1 - \dfrac{1}{\sqrt{2}} + \dfrac{1}{2}\Big)\\[1em] = \big(\dfrac{3}{2} + \dfrac{1}{\sqrt{2}}\Big) \Big(\dfrac{3}{2} - \dfrac{1}{\sqrt{2}}\Big)\\[1em] = \Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{1}{\sqrt{2}}\Big)^2\\[1em] = \dfrac{9}{4} - \dfrac{1}{2}\\[1em] = \dfrac{9 - 2}{4} = \dfrac{7}{4}

Hence, option 3 is the correct option.

Case Study Based Questions

Question 1

Case Study:

One day three friends Amit(A), Binay(B) and Chanchal(C) were playing hide and seek game in the park of their society. Amit and Binay hide in the shrubs and Chanchal has to find both of them. If the position of three friends are at A, B and C respectively, as shown in the figure and forms a right angled triangle ABC such that AB = 6 m, BC = 232\sqrt{3} m and ∠B = 90°.

One day three friends Amit(A), Binay(B) and Chanchal(C) were playing hide and seek game in the park of their society. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Based on the above information answer the following questions:

  1. The length of AC is :
    (a) 4 m
    (b) 838\sqrt{3} m
    (c) 535\sqrt{3} m
    (d) 434\sqrt{3} m

  2. The measure of ∠A is :
    (a) 30°
    (b) 45°
    (c) 60°
    (d) 90°

  3. The measure of ∠C is :
    (a) 30°
    (b) 45°
    (c) 60°
    (d) 90°

  4. cos 2A is equal to :
    (a) 1
    (b) 12\dfrac{1}{2}
    (c) 32\dfrac{\sqrt{3}}{2}
    (d) 3\sqrt{3}

  5. 2sin⁡(C2)2\sin(\dfrac{C}{2}) is equal to :
    (a) 1
    (b) 12\dfrac{1}{2}
    (c) 3\sqrt{3}
    (d) 32\dfrac{\sqrt{3}}{2}

Answer

  1. Perpendicular = BC = 232\sqrt{3} m and Base = AB = 6 m

Hypotenuse = AC

We will find Hypotenuse by using Pythagoras Theorem,

Hypotenuse2 = Perpendicular2 + Base2

Hypotenuse2 = (23)2(2\sqrt{3})^2 + 62

Hypotenuse2 = 12 + 36

Hypotenuse2 = 48

Hypotenuse = 48=43\sqrt{48} = 4\sqrt{3} m

Hence, option (d) is the correct option.

2. sin A = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

sin A = BCAC=2343\dfrac{BC}{AC} = \dfrac{2\sqrt{3}}{4\sqrt{3}}

sin A = 12\dfrac{1}{2}

sin A = sin 30°

A = 30°

Hence, option (a) is the correct option.

3. sin C = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

sin C = ABAC=643\dfrac{AB}{AC} = \dfrac{6}{4\sqrt{3}}

sin C = 32\dfrac{\sqrt{3}}{2}

sin C = sin 60°

C = 60°

Hence, option (c) is the correct option.

4. As, A = 30°

cos 2(30°) = cos 60° = 12\dfrac{1}{2}

Hence, option (b) is the correct option.

5. As, C = 60°

2sin⁡(C2)=2sin⁡(602)2\sin(\dfrac{C}{2}) = 2\sin(\dfrac{60}{2})

= 2 sin 30°

= 2×122 \times \dfrac{1}{2}

= 1.

Hence, option (a) is the correct option.

Question 2

Case Study:

In the figure given below, the rod AB of length 4 inches of a TV disc antena is fixed at right angle to the wall and a rod BC of length 8 inches is supporting the disc.

Based on the above information answer the following questions:

In the figure given below, the rod AB of length 4 inches of a TV disc antena is fixed at right angle to the wall and a rod BC of length 8 inches is supporting the disc. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

  1. The measure of ∠ACB is :
    (a) 30°
    (b) 45°
    (c) 60°
    (d) 90°

  2. The value of tan ∠ABC is :
    (a) 13\dfrac{1}{\sqrt{3}}
    (b) 3\sqrt{3}
    (c) 1
    (d) 0

  3. The value of sin2∠ACB + sin2∠ABC is :
    (a) 12\dfrac{1}{2}
    (b) 0
    (c) 1
    (d) not defined

  4. The length of AC is :
    (a) 6 inches
    (b) 535\sqrt{3} inches
    (c) 424\sqrt{2} inches
    (d) 434\sqrt{3} inches

  5. The value of sin ∠ACB + cos ∠ABC + cot ∠BAC is :
    (a) 0
    (b) 1
    (c) 2
    (d) not defined

Answer

1. AB = Perpendicular = 4 inches

BC = Hypotenuse = 8 inches

sin∠ACB = PerpendicularHypotenuse\dfrac{\text Perpendicular}{\text Hypotenuse}

sin∠ACB = 48=12\dfrac{4}{8} = \dfrac{1}{2}

sin∠ACB = sin 30°

∠ACB = 30°

Hence, option (a) is the correct option.

2. ∠ABC + ∠BAC + ∠ACB = 180°

∠ABC + 90° + 30° = 180°

∠ABC = 180° - 120° = 60°

tan ∠ABC = tan 60° = 3\sqrt{3}.

Hence, option (b) is the correct option.

3. sin2∠ACB + sin2∠ABC

= sin230° + sin260°

= (12)2+(32)2\Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2

= 14+34\dfrac{1}{4} + \dfrac{3}{4}

= 44\dfrac{4}{4}

= 1.

Hence, option (c) is the correct option.

4. AC = Base

By pythagoras theorem,

Hypotenuse2 = Base2 + Perpendicular2

Base2 = Hypotenuse2 - Perpendicular2

AC2 = BC2 - AB2

AC2 = 82 - 42

AC2 = 64 - 16

AC2 = 48

AC = 48\sqrt{48}

AC = 434\sqrt{3} inches.

Hence, option (d) is the correct option.

5. Solving,

sin ∠ACB + cos ∠ABC + cot ∠BAC

= sin 30° + cos 60° + cot 90°

= 12+12+0\dfrac{1}{2} + \dfrac{1}{2} + 0

= 1.

Hence, option (b) is the correct option.

Assertion-Reason Questions

Question 1

Assertion (A): The value of sin 60° cos 30° + cos 60° sin 30° is 0.

Reason (R): sin 90° = 0 and sin 0° = 1.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Given,

⇒sin⁡60∘cos⁡30∘+cos⁡60∘sin⁡30∘⇒32×32+12×12⇒34+14⇒44=1.\Rightarrow \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{3}{4} + \dfrac{1}{4}\\[1em] \Rightarrow \dfrac{4}{4}\\[1em] = 1.

Assertion (A) is false.

As,

sin 90° = 1 and sin 0° = 0.

So,

Reason (R) is false.

Hence, Option 4 is the correct option.

Question 2

Assertion (A): In a right angled △ABC, if ∠ABC = 90°, AB = 3 cm, BC = 4 cm, then sin A = cos C.

Reason (R): sin⁡θcos⁡θ\dfrac{\sin θ}{\cos θ} = tan θ and sin θ ×\times cos θ = cot θ.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

In a right angled △ABC, if ∠ABC = 90, AB = 3 cm, BC = 4 cm, then sin A = cos C. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

In a right angled △ABC,

∠ABC = 90°, AB = 3 cm, BC = 4 cm

By pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

AC2 = (BC)2 + (AB)2

AC2 = 42 + 32

AC2 = 16 + 9

AC2 = 25

AC = 25\sqrt{25}

AC = 5 cm.

⇒sin⁡A=perpendicularhypotenuse=BCAC=45⇒cos⁡C=basehypotenuse=BCAC=45⇒sin⁡A=cos⁡C\Rightarrow \sin A = \dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4}{5}\\[1em] \Rightarrow \cos C = \dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4}{5}\\[1em] \Rightarrow \sin A = \cos C

Assertion (A) is true.

As, sin⁡θcos⁡θ\dfrac{\sin θ}{\cos θ} = tan θ

but sin θ ×\times cos θ is not equal to cot θ.

So, Reason (R) is false.

Hence, Option 1 is the correct option.

Question 3

Assertion (A): sin⁡27∘cos⁡63∘\dfrac{\sin 27^\circ}{\cos 63^\circ} = 1.

Reason (R): sin(90° - θ) = cos θ and cos(90° - θ) = sin θ.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

⇒sin⁡27∘cos⁡63∘=sin⁡(90∘−63∘)cos⁡63∘=cos⁡63∘cos⁡63∘=1\Rightarrow \dfrac{\sin 27^\circ}{\cos 63^\circ}\\[1em] = \dfrac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}\\[1em] = \dfrac{\cos 63^\circ}{\cos 63^\circ}\\[1em] = 1

Assertion (A) is true.

As, sin(90° - θ) = cos θ

and

cos(90° - θ) = sin θ

So,

Reason (R) is true.

Hence, Option 3 is the correct option.

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