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Chapter 8

Introduction to Trigonometry

Class 10 - NCERT Mathematics Solutions



Exercise 8.1

Question 1

In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

(i) sin A, cos A

(ii) sin C, cos C

Answer

ABC is a right angled triangle as shown below:

In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) sin A, cos A (ii) sin C, cos C. NCERT Class 10 Mathematics CBSE Solutions.

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ AC2 = 242 + 72

⇒ AC2 = 576 + 49

⇒ AC2 = 625

⇒ AC = 625\sqrt{625} = 25 cm.

(i) sin A = Side opposite to ∠AHypotenuse=BCAC\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC}

Substituting values we get,

sin A = 725\dfrac{7}{25}.

cos A = Side adjacent to ∠AHypotenuse=ABAC\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC}

Substituting values we get,

cos A = 2425\dfrac{24}{25}.

Hence, sin A = 725\dfrac{7}{25} and cos A = 2425\dfrac{24}{25}.

(ii) sin C = Side opposite to ∠CHypotenuse=ABAC\dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC}

Substituting values we get,

sin C = 2425\dfrac{24}{25}.

cos C = Side adjacent to ∠CHypotenuse=BCAC\dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC}

Substituting values we get,

cos C = 725\dfrac{7}{25}.

Hence, sin C = 2425\dfrac{24}{25} and cos C = 725\dfrac{7}{25}.

Question 2

In the given figure, find tan P – cot R.

In the given figure, find tan P – cot R. NCERT Class 10 Mathematics CBSE Solutions.

Answer

From figure,

PQR is a right angled triangle.

By pythagoras theorem, we get :

⇒ PR2 = PQ2 + QR2

⇒ 132 = 122 + QR2

⇒ QR2 = 169 - 144

⇒ QR2 = 25

⇒ QR = 25\sqrt{25} = 5.

tan P = Side opposite to ∠PSide adjacent to ∠P=QRPQ=512\dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{5}{12}.

cot R = Side adjacent to ∠RSide opposite to ∠R=QRPQ=512\dfrac{\text{Side adjacent to ∠R}}{\text{Side opposite to ∠R}} = \dfrac{QR}{PQ} = \dfrac{5}{12}.

Substituting values in tan P - cot R, we get :

512512\Rightarrow \dfrac{5}{12} - \dfrac{5}{12} = 0.

Hence, tan P - cot R = 0.

Question 3

If sin A = 34\dfrac{3}{4}, calculate cos A and tan A.

Answer

Let us draw a right angle triangle ABC.

If sin A = (3)/(4), calculate cos A and tan A. NCERT Class 10 Mathematics CBSE Solutions.

We know that,

sin A = Side opposite to ∠AHypotenuse\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}}

Substituting values we get,

34=BCAC\dfrac{3}{4} = \dfrac{BC}{AC}

Let BC = 3k and AC = 4k.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ (4k)2 = AB2 + (3k)2

⇒ AB2 = 16k2 - 9k2

⇒ AB2 = 7k2

⇒ AB = 7k2=7k\sqrt{7k^2} = \sqrt{7}k

We know that,

cos A = Side adjacent to ∠AHypotenuse=ABAC=7k4k=74\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{7}k}{4k} = \dfrac{\sqrt{7}}{4}.

tan A = Side opposite to ∠ASide adjacent to ∠A=BCAB=3k7k=37\dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}} = \dfrac{BC}{AB} = \dfrac{3k}{\sqrt{7}k} = \dfrac{3}{\sqrt{7}}.

Hence, cos A=74 and tan A=37\text{cos A} = \dfrac{\sqrt{7}}{4} \text{ and tan A} = \dfrac{3}{\sqrt{7}}.

Question 4

Given 15 cot A = 8, find sin A and sec A.

Answer

Given,

⇒ 15 cot A = 8

⇒ cot A = 815\dfrac{8}{15}

Let us draw a right angle triangle ABC.

Given 15 cot A = 8, find sin A and sec A. NCERT Class 10 Mathematics CBSE Solutions.

We know that,

cot A = Side adjacent to ∠ASide opposite to ∠A\dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}}

Substituting values we get,

815=ABBC\dfrac{8}{15} = \dfrac{AB}{BC}

Let AB = 8k and BC = 15k.

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (8k)2 + (15k)2

⇒ AC2 = 64k2 + 225k2

⇒ AC2 = 289k2

⇒ AC = 289k2\sqrt{289k^2} = 17k.

We know that,

sin A = Side opposite to ∠AHypotenuse=BCAC=15k17k=1517\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{15k}{17k} = \dfrac{15}{17}.

sec A = HypotenuseSide adjacent to ∠A=ACAB=17k8k=178\dfrac{\text{Hypotenuse}}{\text{Side adjacent to ∠A}} = \dfrac{AC}{AB} = \dfrac{17k}{8k} = \dfrac{17}{8}.

Hence, sin A=1517 and sec A=178\text{sin A} = \dfrac{15}{17} \text{ and sec A} = \dfrac{17}{8}.

Question 5

Given sec θ = 1312\dfrac{13}{12}, calculate all other trigonometric ratios.

Answer

Let us draw a right angle triangle ABC, with ∠A = θ.

Given sec θ = (13)/(12), calculate all other trigonometric ratios. NCERT Class 10 Mathematics CBSE Solutions.

We know that,

sec θ = HypotenuseSide adjacent to angle θ\dfrac{\text{Hypotenuse}}{\text{Side adjacent to angle θ}}

Substituting values we get,

1312=ACAB\dfrac{13}{12} = \dfrac{AC}{AB}

Let AC = 13k and AB = 12k.

In right angle triangle ABC,

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ (13k)2 = (12k)2 + BC2

⇒ BC2 = 169k2 - 144k2

⇒ BC2 = 25k2

⇒ BC = 25k2\sqrt{25k^2} = 5k.

We know that,

⇒ sin θ = Side opposite to angle θHypotenuse=BCAC=5k13k=513\dfrac{\text{Side opposite to angle θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{5k}{13k} = \dfrac{5}{13},

⇒ cos θ = 1sec θ=11312=1213\dfrac{1}{\text{sec θ}} = \dfrac{1}{\dfrac{13}{12}} = \dfrac{12}{13}

⇒ tan θ = Side opposite to angle θSide adjacent to angle θ=BCAB=5k12k=512\dfrac{\text{Side opposite to angle θ}}{\text{Side adjacent to angle θ}} = \dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12}.

⇒ cot θ = 1tan θ=1512=125\dfrac{1}{\text{tan θ}} = \dfrac{1}{\dfrac{5}{12}} = \dfrac{12}{5}

⇒ cosec θ = 1sin θ=1513=135\dfrac{1}{\text{sin θ}} = \dfrac{1}{\dfrac{5}{13}} = \dfrac{13}{5}.

Question 6

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Answer

Let us consider two right triangles ACD and BEF where cos A = cos B.

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B. NCERT Class 10 Mathematics CBSE Solutions.

We know that,

cos A = Side adjacent to angle AHypotenuse=ACAD\dfrac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \dfrac{AC}{AD}

cos B = Side adjacent to angle BHypotenuse=BEBF\dfrac{\text{Side adjacent to angle B}}{\text{Hypotenuse}} = \dfrac{BE}{BF}.

Given,

⇒ cos A = cos B

ACAD=BEBF\dfrac{AC}{AD} = \dfrac{BE}{BF}

ACBE=ADBF\dfrac{AC}{BE} = \dfrac{AD}{BF} = k (let) ........(1)

⇒ AC = k.BE and AD = k.BF

In right angle triangle ACD,

By pythagoras theorem,

⇒ AD2 = AC2 + CD2

⇒ CD2 = AD2 - AC2

⇒ CD2 = (k.BF)2 - (k.BE)2

⇒ CD2 = k2(BF2 - BE2)

⇒ CD = k2(BF2BE2)=k(BF2BE2)\sqrt{k^2(BF^2 - BE^2)} = k\sqrt{(BF^2 - BE^2)}

In right angle triangle BEF,

By pythagoras theorem,

⇒ BF2 = BE2 + EF2

⇒ EF2 = BF2 - BE2

⇒ EF = BF2BE2\sqrt{BF^2 - BE^2}

So,

CDEF=kBF2BE2BF2BE2\dfrac{CD}{EF} = \dfrac{k\sqrt{BF^2 - BE^2}}{\sqrt{BF^2 - BE^2}} = k ..........(2)

From (1) and (2), we get :

ACBE=ADBF=CDFE\dfrac{AC}{BE} = \dfrac{AD}{BF} = \dfrac{CD}{FE}

Since, ratio of corresponding sides of similar triangle are proportional.

∴ △ACD ~ △BEF.

∴ ∠A = ∠B.

Hence, proved that ∠A = ∠B.

Question 7

If cot θ = 78\dfrac{7}{8}, evaluate :

(i) (1 + sin θ)(1 - sin θ)(1 + cos θ)(1 - cos θ)\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}}

(ii) cot2 θ

Answer

Let us draw a right angle triangle ABC, with ∠A = θ.

If cot θ = (7)/(8), evaluate : (i) (1 + sin θ)(1 - sin θ)/(1 + cos θ)(1 - cos θ) (ii) cot2 θ. NCERT Class 10 Mathematics CBSE Solutions.

We know that,

cot θ = Side adjacent to ∠θSide opposite to ∠θ\dfrac{\text{Side adjacent to ∠θ}}{\text{Side opposite to ∠θ}}

Substituting values we get :

78=ABBC\dfrac{7}{8} = \dfrac{AB}{BC}

Let AB = 7k and BC = 8k.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC 2

⇒ AC2 = (7k)2 + (8k)2

⇒ AC2 = 49k2 + 64k2

⇒ AC2 = 113k2

⇒ AC = 113k\sqrt{113}k.

(i) We know that,

sin θ = Side opposite to ∠θHypotenuse=BCAC=8k113k=8113\dfrac{\text{Side opposite to ∠θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8k}{\sqrt{113}k} = \dfrac{8}{\sqrt{113}}.

cos θ = Side adjacent to ∠θHypotenuse=ABAC=7k113k=7113\dfrac{\text{Side adjacent to ∠θ}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{7k}{\sqrt{113}k} = \dfrac{7}{\sqrt{113}}.

Substituting values of sin θ and cos θ in (1 + sin θ)(1 - sin θ)(1 + cos θ)(1 - cos θ)\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}}, we get :

(1+8113)(18113)(1+7113)(17113)(1)2(8113)2(1)2(7113)2164113149113113641131134911349113641134964.\Rightarrow \dfrac{\Big(1 + \dfrac{8}{\sqrt{113}}\Big)\Big(1 - \dfrac{8}{\sqrt{113}}\Big)}{\Big(1 + \dfrac{7}{\sqrt{113}}\Big)\Big(1 - \dfrac{7}{\sqrt{113}}\Big)} \\[1em] \Rightarrow \dfrac{(1)^2 - \Big(\dfrac{8}{\sqrt{113}}\Big)^2}{(1)^2 - \Big(\dfrac{7}{\sqrt{113}}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{64}{113}}{1 - \dfrac{49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{113 - 64}{113}}{\dfrac{113 - 49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}} \\[1em] \Rightarrow \dfrac{49}{64}.

Hence, (1 + sin θ)(1 - sin θ)(1 + cos θ)(1 - cos θ)=4964\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}} = \dfrac{49}{64}.

(ii) Given,

⇒ cot θ = 78\dfrac{7}{8}

⇒ cot2 θ = (78)2=4964\Big(\dfrac{7}{8}\Big)^2 = \dfrac{49}{64}.

Hence, cot2 θ = 4964\dfrac{49}{64}.

Question 8

If 3 cot A = 4, check whether 1tan2A1+tan2A\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} = cos2 A - sin2 A or not.

Answer

Given,

⇒ 3 cot A = 4

⇒ cot A = 43\dfrac{4}{3}

Let us draw a right angle triangle ABC.

We know that,

cot A = Side adjacent to ∠ASide opposite to ∠A\dfrac{\text{Side adjacent to ∠A}}{\text{Side opposite to ∠A}}

Substituting values, we get :

ABBC=43\dfrac{\text{AB}}{\text{BC}} = \dfrac{4}{3}

Let AB = 4k and BC = 3k.

If 3 cot A = 4, check whether (1 - (tan)^2 A)/(1 + (tan)^2 A) = cos2 A - sin2 A or not. NCERT Class 10 Mathematics CBSE Solutions.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (4k)2 + (3k)2

⇒ AC2 = 16k2 + 9k2

⇒ AC2 = 25k2

⇒ AC = 25k2\sqrt{25k^2} = 5k.

We know that,

tan A = 1cot A=143=34\dfrac{1}{\text{cot A}} = \dfrac{1}{\dfrac{4}{3}} = \dfrac{3}{4}.

Substituting value of tan A in 1tan2A1+tan2A\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A}

1tan2A1+tan2A1(34)21+(34)219161+9161691616+9167162516725.\Rightarrow \dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} \\[1em] \Rightarrow \dfrac{1 - \Big(\dfrac{3}{4}\Big)^2}{1 + \Big(\dfrac{3}{4}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{9}{16}}{1 + \dfrac{9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{16 - 9}{16}}{\dfrac{16 + 9}{16}} \\[1em] \Rightarrow \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}} \\[1em] \Rightarrow \dfrac{7}{25}.

We know that,

cos A = Side adjacent to ∠AHypotenuse=ABAC=4k5k=45\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{4k}{5k} = \dfrac{4}{5}.

sin A = Side opposite to ∠AHypotenuse=BCAC=3k5k=35\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3k}{5k} = \dfrac{3}{5}.

Substituting value of cos A and sin A in cos2 A - sin2 A, we get :

(45)2(35)21625925725.\Rightarrow \Big(\dfrac{4}{5}\Big)^2 - \Big(\dfrac{3}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{16}{25} - \dfrac{9}{25} \\[1em] \Rightarrow \dfrac{7}{25}.

Hence, proved that 1tan2A1+tan2A\dfrac{1 - \text{tan}^2 A}{1 + \text{tan}^2 A} = cos2 A - sin2 A.

Question 9

In triangle ABC, right-angled at B, if tan A = 13\dfrac{1}{\sqrt{3}}, find the value of :

(i) sin A cos C + cos A sin C

(ii) cos A cos C - sin A sin C

Answer

Let us consider a right angle triangle ABC.

Given,

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

We know that,

tan A = Side opposite to ∠ASide adjacent to ∠A\dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}}

Substituting values, we get :

13=BCAB\dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB}

Let AB = 3\sqrt{3}k and BC = k.

In triangle ABC, right-angled at B, if tan A = (1)/(3), find the value of : (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C. NCERT Class 10 Mathematics CBSE Solutions.

In △ABC,

By pythagoras theorem,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (3k)2+k2(\sqrt{3}k)^2 + k^2

⇒ AC2 = 3k2 + k2

⇒ AC2 = 4k2

⇒ AC = 4k2\sqrt{4k^2} = 2k.

We know that,

sin A = Side opposite to ∠AHypotenuse=BCAC=k2k=12\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}

cos A = Side adjacent to ∠AHypotenuse=ABAC=3k2k=32\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}

sin C = Side opposite to ∠CHypotenuse=ABAC=3k2k=32\dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}

cos C = Side adjacent to ∠CHypotenuse=BCAC=k2k=12\dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}

(i) Substituting values of sin A, cos C, sin C and cos A in sin A cos C + cos A sin C, we get :

12×12+32×3214+34441.\Rightarrow \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{1}{4} + \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{4}{4} \\[1em] \Rightarrow 1.

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

(ii) Substituting values of cos A, cos C, sin A and sin C in cos A cos C - sin A sin C, we get :

32×1212×3234340.\Rightarrow \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} - \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] \Rightarrow 0.

Hence, cos A cos C - sin A sin C = 0.

Question 10

In △ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer

Given,

⇒ PR + QR = 25 cm

⇒ PR = (25 - QR) cm

In △ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. NCERT Class 10 Mathematics CBSE Solutions.

In right angle triangle PQR,

By pythagoras theorem,

⇒ PR2 = PQ2 + QR2

⇒ (25 - QR)2 = 52 + QR2

⇒ 625 + QR2 - 50QR = 25 + QR2

⇒ 50QR = 625 - 25 + QR2 - QR2

⇒ 50QR = 600

⇒ QR = 60050\dfrac{600}{50} = 12

⇒ PR = 25 - QR = 25 - 12 = 13

We know that,

sin P = Side opposite to ∠PHypotenuse=QRPR=1213\dfrac{\text{Side opposite to ∠P}}{\text{Hypotenuse}} = \dfrac{QR}{PR} = \dfrac{12}{13}

cos P = Side adjacent to ∠PHypotenuse=PQPR=513\dfrac{\text{Side adjacent to ∠P}}{\text{Hypotenuse}} = \dfrac{PQ}{PR} = \dfrac{5}{13}

tan P = Side opposite to ∠PSide adjacent to ∠P=QRPQ=125\dfrac{\text{Side opposite to ∠P}}{\text{Side adjacent to ∠P}} = \dfrac{QR}{PQ} = \dfrac{12}{5}

Hence, sin P = 1213, cos P=513 and tan P=125\dfrac{12}{13}, \text{ cos P} = \dfrac{5}{13} \text{ and tan P} = \dfrac{12}{5}.

Question 11

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 125\dfrac{12}{5} for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin θ = 43\dfrac{4}{3} for some angle θ.

Answer

(i) We know that,

tan 45° = 1.

∴ tan A is not always less than 1.

Hence, above statement is false.

(ii) We know that,

sec A ≥ 1

Since, 125\dfrac{12}{5} is greater than 1.

∴ sec A = 125\dfrac{12}{5} is possible.

Hence, above statement is true.

(iii) We know that,

cosine A is also referred as cos A.

Hence, above statement is false.

(iv) We know that,

cot A ≠ cot × A

Hence, above statement is false.

(v) We know that,

0 ≤ sin θ ≤ 1.

Since, 43>1\dfrac{4}{3} \gt 1

∴ sin θ cannot be equal to 43\dfrac{4}{3}.

Hence, above statement is false.

Exercise 8.2

Question 1(i)

Evaluate the following :

sin 60° cos 30° + sin 30° cos 60°

Answer

Substituting values, we get :

sin 60° cos 30° + sin 30° cos 60°=32×32+12×12=34+14=44=1.\Rightarrow \text{sin 60° cos 30° + sin 30° cos 60°} = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1.

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

Question 1(ii)

Evaluate the following :

2 tan2 45° + cos2 30° - sin2 60°

Answer

Substituting values, we get :

2 tan245°+ cos230° sin260°=2×(1)2+(32)2(32)2=2×(1)+3434=2.\Rightarrow \text{2 tan}^2 45° + \text{ cos}^2 30° - \text{ sin}^2 60° = 2 \times (1)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2 \\[1em] = 2 \times (1) + \dfrac{3}{4} - \dfrac{3}{4} \\[1em] = 2.

Hence, 2 tan2 45° + cos2 30° - sin2 60° = 2.

Question 1(iii)

Evaluate the following :

cos 45°sec 30° + cosec 30°\dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}}

Answer

Substituting values, we get :

cos 45°sec 30° + cosec 30°=1223+2=122+233=32(2+23)=322(1+3)=322(1+3)×1313=3(13)22(13)=3(13)22×2=3(13)42=3342=3342×22=3268.\Rightarrow \dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}} = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}} + 2} \\[1em] = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2 + 2\sqrt{3}}{\sqrt{3}}} \\[1em] = \dfrac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} \\[1em] = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \\[1em] = \dfrac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \times \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}} \\[1em] = \dfrac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(1 - 3)} \\[1em] = \dfrac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2} \times -2} \\[1em] = \dfrac{-\sqrt{3}(1 - \sqrt{3})}{4\sqrt{2}} \\[1em] = \dfrac{3 - \sqrt{3}}{4\sqrt{2}} \\[1em] = \dfrac{3 - \sqrt{3}}{4\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\[1em] = \dfrac{3\sqrt{2} - \sqrt{6}}{8}.

Hence, cos 45°sec 30° + cosec 30°=3268.\dfrac{\text{cos 45°}}{\text{sec 30° + cosec 30°}} = \dfrac{3\sqrt{2} - \sqrt{6}}{8}.

Question 1(iv)

Evaluate the following :

sin 30° + tan 45° - cosec 60°sec 30° + cos 60° + cot 45°\dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}}

Answer

Substituting values, we get :

sin 30° + tan 45° - cosec 60°sec 30° + cos 60° + cot 45°=12+12323+12+1=322332+23=322332+23×32233223=(3223)2(32)2(23)2=(32)2+(23)22×32×23(32)2(23)2=94+43239443=27+1624312271612=4324311.\Rightarrow \dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}} = \dfrac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1} \\[1em] = \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} + \dfrac{2}{\sqrt{3}}} \\[1em] = \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} + \dfrac{2}{\sqrt{3}}} \times \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}} \\[1em] = \dfrac{\Big(\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}\Big)^2}{\Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{2}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\Big(\dfrac{3}{2}\Big)^2 + \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - 2 \times \dfrac{3}{2} \times \dfrac{2}{\sqrt{3}}}{\Big(\dfrac{3}{2}\Big)^2 - \Big(\dfrac{2}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{9}{4} + \dfrac{4}{3} - 2\sqrt{3}}{\dfrac{9}{4} - \dfrac{4}{3}} \\[1em] = \dfrac{\dfrac{27 + 16 - 24\sqrt{3}}{12}}{\dfrac{27 - 16}{12}} \\[1em] = \dfrac{43 - 24\sqrt{3}}{11}.

Hence, sin 30° + tan 45° - cosec 60°sec 30° + cos 60° + cot 45°=4324311.\dfrac{\text{sin 30° + tan 45° - cosec 60°}}{\text{sec 30° + cos 60° + cot 45°}} = \dfrac{43 - 24\sqrt{3}}{11}.

Question 1(v)

Evaluate the following :

5 cos260°+4 sec230° tan245°sin230°+ cos230°\dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°}

Answer

Solving,

5 cos260°+4 sec230° tan245°sin230°+ cos230°5×(12)2+4×(23)212(12)2+(32)25×14+4×43114+3454+16314415+64121216712.\Rightarrow \dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} \\[1em] \Rightarrow \dfrac{5 \times \Big(\dfrac{1}{2}\Big)^2 + 4 \times \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - 1^2}{\Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2} \\[1em] \Rightarrow \dfrac{5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1}{\dfrac{1}{4} + \dfrac{3}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{5}{4} + \dfrac{16}{3} - 1}{\dfrac{4}{4}} \\[1em] \Rightarrow \dfrac{\dfrac{15 + 64 - 12}{12}}{1} \\[1em] \Rightarrow \dfrac{67}{12}.

Hence, 5 cos260°+4 sec230° tan245°sin230°+ cos230°=6712\dfrac{5\text{ cos}^2 60° + 4\text{ sec}^2 30° - \text{ tan}^2 45°}{\text{sin}^2 30° + \text{ cos}^2 30°} = \dfrac{67}{12}.

Question 2(i)

Choose the correct option and justify your choice :

2 tan 30°1 + tan230°=\dfrac{\text{2 tan 30°}}{\text{1 + tan}^2 30°} =

  1. sin 60°

  2. cos 60°

  3. tan 60°

  4. sin 30°

Answer

Solving,

2 tan 30°1 + tan230°=2×131+(13)2=231+13=233+13=2343=2×34×3=32=sin 60°.\Rightarrow \dfrac{\text{2 tan 30°}}{\text{1 + tan}^2 30°} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}} \\[1em] = \dfrac{2 \times 3}{4 \times \sqrt{3}} \\[1em] = \dfrac{\sqrt{3}}{2} \\[1em] = \text{sin 60°}.

Hence, Option 1 is the correct option.

Question 2(ii)

Choose the correct option and justify your choice :

1 - tan245°1 + tan245°=\dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} =

  1. tan 90°

  2. 1

  3. sin 45°

  4. 0

Answer

Solving,

1 - tan245°1 + tan245°=111+1=02=0.\Rightarrow \dfrac{\text{1 - tan}^2 45°}{\text{1 + tan}^2 45°} = \dfrac{1 - 1}{1 + 1} \\[1em] = \dfrac{0}{2} \\[1em] = 0.

Hence, Option 4 is the correct option.

Question 2(iii)

Choose the correct option and justify your choice :

sin 2A = 2 sin A is true when A =

  1. 30°

  2. 45°

  3. 60°

Answer

Given,

Equation : sin 2A = 2 sin A

Substituting A = 0° in L.H.S. of above equation :

⇒ sin 2(0°) = sin 0° = 0.

Substituting A = 0° in R.H.S. of above equation :

⇒ 2 sin 0° = 2 sin 0° = 0.

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

Hence, Option 1 is the correct option.

Question 2(iv)

Choose the correct option and justify your choice :

2 tan 30°1 - tan230°=\dfrac{\text{2 tan 30°}}{\text{1 - tan}^2 30°} =

  1. cos 60°

  2. sin 60°

  3. tan 60°

  4. sin 30°

Answer

Solving,

2 tan 30°1 - tan230°=2×131(13)2=23113=2323=2×32×3=3=tan 60°.\Rightarrow \dfrac{\text{2 tan 30°}}{\text{1 - tan}^2 30°} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\[1em] = \dfrac{2 \times 3}{2 \times \sqrt{3}} \\[1em] = \sqrt{3} \\[1em] = \text{tan 60°}.

Hence, Option 3 is the correct option.

Question 3

If tan (A + B) = 3\sqrt{3} and tan (A – B) = 13\dfrac{1}{\sqrt{3}}; 0° < A + B ≤ 90°; A > B, find A and B.

Answer

Given,

⇒ tan (A + B) = 3\sqrt{3}

⇒ tan (A + B) = tan 60°

⇒ A + B = 60° ..........(1)

Also,

⇒ tan (A - B) = 13\dfrac{1}{\sqrt{3}}

⇒ tan (A - B) = tan 30°

⇒ A - B = 30° ..........(2)

Adding equation (1) and (2), we get :

⇒ A + B + A - B = 60° + 30°

⇒ 2A = 90°

⇒ A = 90°2\dfrac{90°}{2} = 45°.

Substituting value of A in equation (1), we get :

⇒ 45° + B = 60°

⇒ B = 60° - 45° = 15°.

Hence, A = 45° and B = 15°.

Question 4

State whether the following are true or false. Justify your answer.

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

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Answer

(i) Let A = 30° and B = 60°.

Substituting value of A and B in L.H.S. of the equation sin (A + B) = sin A + sin B, we get :

⇒ sin(A + B)

⇒ sin(30° + 60°)

⇒ sin 90°

⇒ 1.

Substituting value of A and B in R.H.S. of the equation sin (A + B) = sin A + sin B, we get :

⇒ sin A + sin B

⇒ sin 30° + sin 60°

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

1+32\dfrac{1 + \sqrt{3}}{2}.

Since, sin (A + B) ≠ sin A + sin B.

Hence, statement sin (A + B) = sin A + sin B is false.

(ii) Let θ be 0°, 30° and 45°.

sin 0° = 0, sin 30° = 12\dfrac{1}{2} and sin 45° = 12\dfrac{1}{\sqrt{2}}.

We see that,

sin θ increases as θ increases.

Hence, the statement the value of sin θ increases as θ increases is true.

(iii) Let θ be 0°, 30° and 45°.

cos 0° = 1, cos 30° = 32\dfrac{\sqrt{3}}{2} and cos 45° = 12\dfrac{1}{\sqrt{2}}.

We see that,

cos θ decreases as θ increases.

Hence, the statement the value of cos θ increases as θ increases is false.

(iv) Let θ be 0°, 30° and 45°.

sin 0° = 0, sin 30° = 12\dfrac{1}{2} and sin 45° = 12\dfrac{1}{\sqrt{2}},

cos 0° = 1, cos 30° = 32\dfrac{\sqrt{3}}{2} and sin 45° = 12\dfrac{1}{\sqrt{2}},

From above we see that,

sin θ = cos θ, only when θ = 45°.

Hence, the statement sin θ = cos θ for all values of θ is false.

(v) We know that,

cot A = cos Asin A\dfrac{\text{cos A}}{\text{sin A}}

For A = 0°, sin A = 0.

∴ cot A is not defined.

Hence, the statement cot A is not defined for A = 0° is true.

Exercise 8.3

Question 1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer

Converting sin A in terms of cot A,

sin Asin2Asin2A1sin2Asin2A+ cos2A1sin2Asin2A+cos2Asin2A11+cot2A.\Rightarrow \text{sin A} \\[1em] \Rightarrow \sqrt{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{1} \\[1em] \Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{\sqrt{\text{sin}^2 A + \text{ cos}^2 A}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{\dfrac{\text{sin}^2 A}{\text{sin}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{1 + \text{cot}^2 A}}.

Converting sec A in terms of cot A,

We know that,

sec2A=1 + tan2Asec2A=1+1cot2Asec2A=cot2A+1cot2Asec A=cot2A+1cot2Asec A=1+cot2Acot A.\Rightarrow \text{sec}^2 A = \text{1 + tan}^2 A \\[1em] \Rightarrow \text{sec}^2 A = 1 + \dfrac{1}{\text{cot}^2 A}\\[1em] \Rightarrow \text{sec}^2 A = \dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A} \\[1em] \Rightarrow \text{sec A} = \sqrt{\dfrac{\text{cot}^2 A + 1}{\text{cot}^2 A}} \\[1em] \Rightarrow \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}}.

We know that,

tan A = 1cot A\dfrac{1}{\text{cot A}}

Hence, sin A = 11+cot2A,sec A=1+cot2Acot A and tan A=1cot A.\dfrac{1}{\sqrt{1 + \text{cot}^2 A}}, \text{sec A} = \dfrac{\sqrt{1 + \text{cot}^2 A}}{\text{cot A}} \text{ and tan A} = \dfrac{1}{\text{cot A}}.

Question 2

Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer

We know that,

⇒ cos A = 1sec A\dfrac{1}{\text{sec A}} .........(1)

By formula,

⇒ sin2 A + cos2 A = 1

⇒ sin2 A = 1 - cos2 A

⇒ sin2 A = 1 - 1sec2A\dfrac{1}{\text{sec}^2 A}

⇒ sin2 A = sec2A1sec2A\dfrac{\text{sec}^2 A - 1}{\text{sec}^2 A}

⇒ sin A = sec2A1sec A\dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}}

By formula,

⇒ sec2 A - tan2 A = 1

⇒ tan2 A = sec2 A - 1

⇒ tan A = sec2A1\sqrt{\text{sec}^2 A - 1}

By formula,

⇒ cot A = 1tan A=1sec2A1\dfrac{1}{\text{tan A}} = \dfrac{1}{\sqrt{\text{sec}^2 A - 1}}

By formula,

⇒ cosec A = 1sin A=1sec2A1sec A=sec Asec2A1\dfrac{1}{\text{sin A}} = \dfrac{1}{\dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}}} = \dfrac{\text{sec A}}{\sqrt{\text{sec}^2 A - 1}}.

Hence,

sin A=sec2A1sec A,cos A=1sec A,tan A=sec2A1, cot A=1sec2A1,cosec A=sec Asec2A1\text{sin A} = \dfrac{\sqrt{\text{sec}^2 A - 1}}{\text{sec A}}, \text{cos A} = \dfrac{1}{\text{sec A}}, \text{tan A} = \sqrt{\text{sec}^2 A - 1}, \text{ cot A} = \dfrac{1}{\sqrt{\text{sec}^2 A - 1}}, \text{cosec A} = \dfrac{\text{sec A}}{\sqrt{\text{sec}^2 A - 1}}

Question 3(i)

Choose the correct option. Justify your choice.

9 sec2 A - 9 tan2 A =

  1. 1

  2. 9

  3. 8

  4. 0

Answer

Solving,

⇒ 9 sec2 A - 9 tan2 A

⇒ 9 (sec2 A - tan2 A)

By formula,

sec2 A - tan2 = 1

⇒ 9 × 1 = 9.

Hence, Option 2 is the correct option.

Question 3(ii)

Choose the correct option. Justify your choice.

(1 + tan θ + sec θ)(1 + cot θ - cosec θ) =

  1. 0

  2. 1

  3. 2

  4. -1

Answer

Solving,

(1 + tan θ + sec θ)(1 + cot θ - cosec θ)(1+sin θcos θ+1cos θ)(1+cos θsin θ1sin θ)(sin θ + cos θ + 1cos θ)(sin θ + cos θ - 1sin θ)(sin θ + cos θ)212sin θ. cos θsin2θ+cos2θ+2 sin θ. cos θ - 1sin θ. cos θ1+2 sin θ.cos θ - 1sin θ. cos θ2 sin θ. cos θsin θ.cos θ2.\Rightarrow \text{(1 + tan θ + sec θ)(1 + cot θ - cosec θ)} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{1}{\text{cos θ}}\Big)\Big(1 + \dfrac{\text{cos θ}}{\text{sin θ}} - \dfrac{1}{\text{sin θ}}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin θ + cos θ + 1}}{\text{cos θ}}\Big)\Big(\dfrac{\text{sin θ + cos θ - 1}}{\text{sin θ}}\Big) \\[1em] \Rightarrow \dfrac{\text{(sin θ + cos θ)}^2 - 1^2}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ + \text{cos}^2 θ + \text{2 sin θ. cos θ - 1}}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{1 + \text{2 sin θ.cos θ - 1}}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{2 sin θ. cos θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow 2.

Hence, Option 3 is the correct option.

Question 3(iii)

Choose the correct option. Justify your choice.

(sec A + tan A) (1 – sin A) =

  1. sec A

  2. sin A

  3. cosec A

  4. cos A

Answer

Solving,

(sec A + tan A) (1 – sin A)(1cos A+sin Acos A)(1 - sin A)(1 + sin Acos A)(1 - sin A)1 - sin2Acos Acos2Acos Acos A.\Rightarrow \text{(sec A + tan A) (1 – sin A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big)\text{(1 - sin A)} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A}.

Hence, Option 4 is the correct option.

Question 3(iv)

Choose the correct option. Justify your choice.

1 + tan2A1 + cot2A=\dfrac{\text{1 + tan}^2 A}{\text{1 + cot}^2 A} =

  1. sec2 A

  2. -1

  3. cot2 A

  4. tan2 A

Answer

Solving,

1 + tan2A1 + cot2Asec2Acosec2A1cos2A1sin2Asin2Acos2Atan2A.\Rightarrow \dfrac{\text{1 + tan}^2 A}{\text{1 + cot}^2 A} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A}{\text{cosec}^2 A} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos}^2 A}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{tan}^2 A.

Hence, Option 4 is the correct option.

Question 4(i)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

(cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}

Answer

Given,

Equation : (cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}

Solving L.H.S. of the above equation :

(1sin θcos θsin θ)2(1cos θsin θ)2(1 - cos θ)2sin2θ(1 - cos θ)21cos2θ(1 - cos θ)2(1 - cos θ)(1 + cos θ)1 - cos θ1 + cos θ.\Rightarrow \Big(\dfrac{1}{\text{sin θ}} - \dfrac{\text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \text{cos θ}}{\text{sin θ}}\Big)^2 \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{sin}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{1 - \text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{(\text{1 - cos θ})^2}{\text{(1 - cos θ)(1 + cos θ)}} \\[1em] \Rightarrow \dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}.

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

Hence, proved that (cosec θ - cot θ)2 = 1 - cos θ1 + cos θ\dfrac{\text{1 - cos θ}}{\text{1 + cos θ}}.

Question 4(ii)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

cos A1 + sin A+1 + sin Acos A\dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} = 2 sec A.

Answer

Given,

Equation : cos A1 + sin A+1 + sin Acos A\dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} = 2 sec A.

Solving L.H.S. of the equation :

cos A1 + sin A+1 + sin Acos Acos2A+(1 + sin A)2cos A(1 + sin A)cos2A+1+sin2A+2 sin Acos A(1 + sin A)sin2A+cos2A+1+2 sin Acos A(1 + sin A)1+1+2 sin Acos A(1 + sin A)2+2 sin Acos A(1 + sin A)2(1+sin A)cos A(1 + sin A)2cos A2 sec A.\Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + \text{(1 + sin A)}^2}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + 1 + \text{sin}^2 A + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 1 + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1 + 1 + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2 + \text{2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2(1 + \text{sin A})}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2}{\text{cos A}} \\[1em] \Rightarrow \text{2 sec A}.

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

Hence, proved that cos A1 + sin A+1 + sin Acos A\dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{1 + sin A}}{\text{cos A}} = 2 sec A.

Question 4(iii)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

tan θ1 - cot θ+cot θ1 - tan θ\dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}} = 1 + sec θ cosec θ

Answer

Solving,

sin θcos θ1cos θsin θ+cos θsin θ1sin θcos θsin θcos θsin θ - cos θsin θ+cos θsin θcos θ - sin θcos θsin2θcos θ(sin θ - cos θ)+cos2θsin θ(cos θ - sin θ)sin2θcos θ(sin θ - cos θ)cos2θsin θ(sin θ - cos θ)sin3θcos3θsin θ cos θ (sin θ - cos θ)(sin θ - cos θ)(sin2θ+cos2θ+sin θ cos θ)sin θ cos θ (sin θ - cos θ)(sin2θ+cos2θ+sin θ cos θ)sin θ cos θ1+sin θ cos θsin θ cos θ1sin θ cos θ+sin θ cos θsin θ cos θsec θ cosec θ+1.\Rightarrow \dfrac{\dfrac{\text{sin θ}}{\text{cos θ}}}{1 - \dfrac{\text{cos θ}}{\text{sin θ}}} + \dfrac{\dfrac{\text{cos θ}}{\text{sin θ}}}{1 - \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{\text{sin θ - cos θ}}{\text{sin θ}}} + \dfrac{\dfrac{\text{cos θ}}{\text{sin θ}}}{\dfrac{\text{cos θ - sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ}{\text{cos θ(sin θ - cos θ)}} + \dfrac{\text{cos}^2 θ}{\text{sin θ(cos θ - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ}{\text{cos θ(sin θ - cos θ)}} - \dfrac{\text{cos}^2 θ}{\text{sin θ(sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{\text{sin}^3 θ - \text{cos}^3 θ}{\text{sin θ cos θ (sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{\text{(sin θ - cos θ)}(\text{sin}^2 θ + \text{cos}^2 θ + \text{sin θ cos θ})}{\text{sin θ cos θ (sin θ - cos θ)}} \\[1em] \Rightarrow \dfrac{(\text{sin}^2 θ + \text{cos}^2 θ + \text{sin θ cos θ})}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1 + \text{sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{1}{\text{sin θ cos θ}} + \dfrac{\text{sin θ cos θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \text{sec θ cosec θ} + 1.

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

Hence, proved that tan θ1 - cot θ+cot θ1 - tan θ\dfrac{\text{tan θ}}{\text{1 - cot θ}} + \dfrac{\text{cot θ}}{\text{1 - tan θ}} = 1 + sec θ cosec θ.

Question 4(iv)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

1 + sec Asec A=sin2A1 - cos A\dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}}

Answer

To prove:

1 + sec Asec A=sin2A1 - cos A\dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}}

Solving L.H.S. of the equation :

1+1cos A1cos Acos A + 1cos A1cos A1 + cos A.\Rightarrow \dfrac{1 + \dfrac{1}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos A + 1}}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \text{1 + cos A}.

Solving R.H.S. of the equation :

sin2A1 - cos A1cos2A1 - cos A(1 - cos A)(1 + cos A)1 - cos A1 + cos A.\Rightarrow \dfrac{\text{sin}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{cos}^2 A}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{1 - cos A}} \\[1em] \Rightarrow \text{1 + cos A}.

Since, L.H.S. = R.H.S. = 1 + cos A.

Hence, proved that 1 + sec Asec A=sin2A1 - cos A\dfrac{\text{1 + sec A}}{\text{sec A}} = \dfrac{\text{sin}^2 A}{\text{1 - cos A}}.

Question 4(v)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

cos A - sin A + 1cos A + sin A - 1\dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A.

Answer

To prove:

cos A - sin A + 1cos A + sin A - 1\dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} = cosec A + cot A

We know that,

⇒ cosec2 A = 1 + cot2 A

⇒ 1 = cosec2 A - cot2 A .....(1)

Dividing numerator and denominator of L.H.S. of the given equation with sin A, we get :

cos A - sin A + 1cos A + sin A - 1cos Asin Asin Asin A+1sin Acos Asin A+sin Asin A1sin Acot A - 1 + cosec Acot A + 1 - cosec A\Rightarrow \dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{cos A}}{\text{sin A}} - \dfrac{\text{sin A}}{\text{sin A}} + \dfrac{1}{\text{sin A}}}{\dfrac{\text{cos A}}{\text{sin A}} + \dfrac{\text{sin A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{cot A - 1 + cosec A}}{\text{cot A + 1 - cosec A}}

Substituting value of 1 from equation (1) in numerator of above equation, we get :

cot A + cosec A(cosec2Acot2A)cot A + 1 - cosec Acot A + cosec A(cosecAcotA)(cosec A + cot A)cot A + 1 - cosec A(cosec A + cot A)[1 - (cosec A - cot A)]cot A - cosec A + 1(cosec A + cot A)(cot A - cosec A + 1)cot A - cosec A + 1cosec A + cot A.\Rightarrow \dfrac{\text{cot A + cosec A} - \text{(cosec}^2 A - \text{cot}^2 A)}{\text{cot A + 1 - cosec A}} \\[1em] \Rightarrow \dfrac{\text{cot A + cosec A} - \text{(cosec} A - \text{cot} A)\text{(cosec A + cot A)}}{\text{cot A + 1 - cosec A}} \\[1em] \Rightarrow \dfrac{\text{(cosec A + cot A)[1 - (cosec A - cot A)]}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \dfrac{\text{(cosec A + cot A)(cot A - cosec A + 1)}}{\text{cot A - cosec A + 1}} \\[1em] \Rightarrow \text{cosec A + cot A}.

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

Hence, proved that cos A - sin A + 1cos A + sin A - 1\dfrac{\text{cos A - sin A + 1}}{\text{cos A + sin A - 1}} = cosec A + cot A.

Question 4(vi)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

1 + sin A1 - sin A\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = sec A + tan A

Answer

To prove:

1 + sin A1 - sin A\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = sec A + tan A.

Solving L.H.S. of the above equation, we get :

1 + sin A1 - sin A×1 + sin A1 + sin A(1 + sin A)21 - sin2A(1 + sin A)2cos2A1 + sin Acos A1cos A+sin Acos Asec A + tan A.\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} \times \sqrt{\dfrac{\text{1 + sin A}}{\text{1 + sin A}}} \\[1em] \Rightarrow \sqrt{\dfrac{(\text{1 + sin A})^2}{\text{1 - sin}^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{(\text{1 + sin A})^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}.

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

Hence, proved that 1 + sin A1 - sin A\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = sec A + tan A.

Question 4(vii)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

sin θ - 2 sin3θ2 cos3θcos θ\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} = tan θ

Answer

To prove:

sin θ - 2 sin3θ2 cos3θcos θ\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} = tan θ

Solving L.H.S. of the above equation :

sin θ(1 - 2 sin2θ)cos θ (2 cos2θ1)sin θ[1 - 2 (1 - cos2θ)]cos θ (2 cos2θ1)sin θ[1 - 2 + 2cos2θ)]cos θ (2 cos2θ1)tan θ (2 cos2θ1)2 cos2θ1tan θ.\Rightarrow \dfrac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 (1 - cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{sin θ[1 - 2 + 2cos}^2 θ)]}{\text{cos θ (2 cos}^2 θ - 1)} \\[1em] \Rightarrow \dfrac{\text{tan θ}\text{ (2 cos}^2 θ - 1)}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}.

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

Hence, proved that sin θ - 2 sin3θ2 cos3θcos θ\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{cos θ}} = tan θ.

Question 4(viii)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Answer

To prove:

(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Solving L.H.S. of the above equation :

⇒ (sin A + cosec A)2 + (cos A + sec A)2

⇒ sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A

⇒ sin2 A + cos2 A + cosec2 A + sec2 A + 2 sin A ×1sin A+ 2 cos A×1cos A\times \dfrac{1}{\text{sin A}} + \text{ 2 cos A} \times \dfrac{1}{\text{cos A}}

⇒ 1 + cosec2 A + sec2 A + 2 + 2

⇒ 5 + (1 + cot2 A) + (1 + tan2 A)

⇒ 7 + tan2 A + cot2 A.

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

Hence, proved that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.

Question 4(ix)

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.

(cosec A – sin A)(sec A – cos A) = 1tan A + cot A\dfrac{1}{\text{tan A + cot A}}