Trigonometrical Ratios — Exercise 21(B)

If $\text{cos A} = \dfrac{1}{\sqrt2}$ and $\text{sin B} = \dfrac{\sqrt3}{2}$, the value of $\dfrac{\text{tan B} - \text{tan A}}{1 + \text{tan A tan B}}$ is :

1. $2 - {\sqrt3}$
2. $2 + {\sqrt3}$
3. ${\sqrt3} - 2$
4. ${\sqrt3} + 2$

Answer

Given:

cos $A = \dfrac{1}{\sqrt2}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt2}$

∴ If length of AM = x unit, length of AO = x $\sqrt{2}$ unit.

In Δ AMO,

⇒ AO² = AM² + MO² (∵ AC is hypotenuse)

⇒ ($\sqrt{2}$x)² = (x)² + MO²

⇒ 2x² = x² + MO²

⇒ MO² = 2x² - x²

⇒ MO² = x²

⇒ MO = $\sqrt{\text{x}^2}$

⇒ MO = x

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{OM}{MA} = \dfrac{x}{x} = 1$

And,

sin $B = \dfrac{\sqrt3}{2}$

i.e. $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{\sqrt3}{2}$

∴ If length of XY = $\sqrt{3}$ y unit, length of YB = 2y unit.

In Δ BXY,

⇒ YB² = YX² + BX² (∵ AC is hypotenuse)

⇒ (2y)² = ($\sqrt{3}$y)² + BX²

⇒ 4y² = 3y² + BX²

⇒ BX² = 4y² - 3y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

tan $B = \dfrac{Perpendicular}{Base}$

$= \dfrac{XY}{BX} = \dfrac{\sqrt{3}y}{y} = \dfrac{\sqrt{3}}{1}$

Now,

$\dfrac{\text{tan B} - \text{tan A}}{1 + \text{tan A } \text{tan B}}\\[1em] = \dfrac{\dfrac{\sqrt{3}}{1} - 1}{1 + \dfrac{\sqrt{3}}{1} \times 1}\\[1em] = \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}}\\[1em] = \dfrac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \dfrac{(\sqrt{3} - 3 - 1 + \sqrt{3})}{1^2 - \sqrt{3}^2}\\[1em] = \dfrac{2 \sqrt{3} - 4}{1 - 3}\\[1em] = \dfrac{2 (\sqrt{3} - 2)}{-2}\\[1em] = \dfrac{\cancel{2} (\sqrt{3} - 2)}{-\cancel{2}}\\[1em] = -\sqrt{3} + 2\\[1em] = 2 - \sqrt{3}$

Hence, option 1 is the correct option.

From the given figure, the value of cos y is :

1. $\dfrac{1}{3}$
2. $\dfrac{1}{4}$
3. $\dfrac{12}{13}$
4. $1\dfrac{1}{12}$

Answer

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

In Δ ADC,

⇒ DC² = AC² + DA² (∵ DC is hypotenuse)

⇒ DC² = 5² + 12²

⇒ DC² = 25 + 144

⇒ DC² = 169

⇒ DC = $\sqrt{169}$

⇒ DC = 13

cos y = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{DA}{DC}\\[1em] = \dfrac{12}{13}$

Hence, option 3 is the correct option.

ABCD is a rhombus with diagonals BD and AC equal to 12 cm and 16 cm respectively. The value of cosec x is :

1. $1\dfrac{2}{3}$
2. $3\dfrac{1}{3}$
3. $\dfrac{1}{2}$
4. $1\dfrac{1}{3}$

Answer

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

In Δ AOB,

⇒ AB² = BO² + OA² (∵ AB is hypotenuse)

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

cosec $x = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AB}{OB}\\[1em] = \dfrac{10}{6}\\[1em] = \dfrac{5}{3}\\[1em] = 1\dfrac{2}{3}$

Hence, option 1 is the correct option.

If sin A - cosec A = 2, the value of sin² A + cosec² A is :

1. 2
2. 0
3. 4
4. 6

Answer

Given:

sin A - cosec A = 2

Squaring both sides,

⇒ (sin A - cosec A)² = (2)²

⇒ (sin A)² + (cosec A)² - 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A - 2 x $\text{sin A}$ x $\dfrac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A - 2 x $\cancel{\text{sin A}}$x $\dfrac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A - 2 = 4

⇒ sin² A + cosec² A = 4 + 2

⇒ sin² A + cosec² A = 6

Hence, option 4 is the correct option.

If $\text{cot A} = {\sqrt5}$, the value of cosec² A - sec² A is :

1. $\dfrac{5}{24}$
2. $4\dfrac{4}{5}$
3. 5
4. 24

Answer

Given:

$\text{cot A} = {\sqrt5}\\[1em] \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{\sqrt5}{1}\\[1em]$

∴ If length of AB = $\sqrt{5}$ x unit, length of BC = x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = $(\sqrt{5}$ x)² + (x)²

⇒ AC² = 5x² + x²

⇒ AC² = 6x²

⇒ AC = $\sqrt{6\text{x}^2}$

⇒ AC = $\sqrt{6}$ x

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{\sqrt{6}\text{x}}{\text{x}} = \dfrac{\sqrt{6}}{1}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{\sqrt{6}\text{x}}{\sqrt{5}\text{x}} = \dfrac{\sqrt{6}}{\sqrt{5}}$

Now, cosec² A - sec² A

$= \Big(\dfrac{\sqrt{6}}{1}\Big)^2 - \Big(\dfrac{\sqrt{6}}{\sqrt{5}}\Big)^2\\[1em] = \dfrac{6}{1} - \dfrac{6}{5}\\[1em] = \dfrac{6 \times 5}{1 \times 5} - \dfrac{6 \times 1}{5 \times 1}\\[1em] = \dfrac{30}{5} - \dfrac{6}{5}\\[1em] = \dfrac{30 - 6}{5}\\[1em] = \dfrac{24}{5}\\[1em] = 4\dfrac{4}{5}$

Hence, option 2 is the correct option.

From the following figure, find :

(i) y

(ii) sin x°

(iii) (sec x° - tan x°)(sec x° + tan x°)

Answer

(i) In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ 2² = y² + 1²

⇒ 4 = y² + 1

⇒ y² = 4 - 1

⇒ y² = 3

⇒ y = $\sqrt{3}$

Hence, the value of y = $\sqrt{3}$.

(ii) sin $x° = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{\sqrt{3}}{2}$

Hence, sin $x° = \dfrac{\sqrt{3}}{2}$.

(iii) (sec x° - tan x°)(sec x° + tan x°)

sec $x° = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{CB}\\[1em] = \dfrac{2}{1} = 2$

tan $x° = \dfrac{Perpendicular}{Base}$

$= \dfrac{AB}{CB}\\[1em] = \dfrac{\sqrt{3}}{1} = 3$

Now, (sec x° - tan x°)(sec x° + tan x°)

$= (2 - \sqrt{3})(2 + \sqrt{3}) \\[1em] = (2)^2 - (\sqrt{3})^2 \\[1em] = 4 - 3 \\[1em] = 1$

Hence, (sec x° - tan x°)(sec x° + tan x°) = 1.

Use the given figure to find :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

Answer

In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ BD² = 6² + 8²

⇒ BD² = 36 + 64

⇒ BD² = 100

⇒ BD = $\sqrt{100}$

⇒ BD = 10

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ 17² = AC² + 8²

⇒ 289 = AC² + 64

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) sin $x° = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{DC}{AD}\\[1em] = \dfrac{8}{17}$

Hence, sin $x° = \dfrac{8}{17}$.

(ii) cos $y° = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Hence, cos $y° = \dfrac{3}{5}$.

(iii) 3 tan x° - 2 sin y° + 4 cos y°

tan $x° = \dfrac{Perpendicular}{Base}$

$= \dfrac{DC}{AC}\\[1em] = \dfrac{8}{15}$

sin $y° = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{CD}{BD}\\[1em] = \dfrac{8}{10}\\[1em] = \dfrac{4}{5}$

cos $y° = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Now, 3 tan x° - 2 sin y° + 4 cos y°

$= 3 \times \dfrac{8}{15} - 2 \times \dfrac{4}{5} + 4 \times \dfrac{3}{5}\\[1em] = \dfrac{24}{15} - \dfrac{8}{5} + \dfrac{12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{- 8 + 12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4 \times 3}{5 \times 3}\\[1em] = \dfrac{24}{15} + \dfrac{12}{15}\\[1em] = \dfrac{24 + 12}{15}\\[1em] = \dfrac{36}{15}\\[1em] = \dfrac{12}{5}\\[1em] = 2\dfrac{2}{5}$

Hence, 3 tan x° - 2 sin y° + 4 cos y° = $2\dfrac{2}{5}$.

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find :

(i) cos ∠DBC

(ii) cot ∠DBA

Answer

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$

⇒ AC = 13

Let ∠CBD be x°.

So, ∠DBA = 90° - x°.

In Δ DAB, according to angle sum property

⇒ ∠ DAB + ∠ADB + ∠DBA = 180°

⇒ ∠ DAB + 90° + (90° - x°) = 180°

⇒ ∠ DAB + 180° - x° = 180°

⇒ ∠ DAB - x° = 0

⇒ ∠ DAB = x°

From figure,

∠ DAB = ∠ CAB = x°

∴ ∠ CBD = ∠ CAB = x°

(i) cos ∠DBC = cos ∠CAB = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{12}{13}$

Hence, cos ∠DBC = $\dfrac{12}{13}$.

(ii) In Δ BCD, according to angle sum property

⇒ ∠ DBC + ∠DCB + ∠CDB = 180°

⇒ ∠ DCB + x° + 90° = 180°

⇒ ∠ DCB = 180° - 90° - x°

⇒ ∠ DCB = 90° - x°

From figure,

∠ DCB = ∠ ACB = 90° - x°

∴ ∠ DBA = ∠ ACB = 90° - x°

cot ∠DBA = cot ∠ACB = $\dfrac{Base}{Perpendicular}$

$= \dfrac{BC}{AB}\\[1em] = \dfrac{5}{12}$

Hence, cot ∠DBA = $\dfrac{5}{12}$.

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Find :

(i) tan ∠DBC

(ii) sin ∠DBA

Answer

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

Let CD = y and BD = x

In Δ BCD,

⇒ BC² = CD² + BD² (∵ BC is hypotenuse)

⇒ 3² = y² + x²

⇒ 9 = y² + x² ................(1)

In Δ ABD,

⇒ AB² = AD² + BD² (∵ BC is hypotenuse)

⇒ 4² = (5 - y)² + x²

⇒ 16 = 25 + y² - 10y + x² ................(2)

Subtracting (1) from (2), we get

⇒ 16 - 9 = (25 + y² - 10y + x²) - (y² + x²)

⇒ 7 = 25 + y² - 10y + x² - y² - x²

⇒ 7 = 25 - 10y

⇒ 7 - 25 = -10y

⇒ -10y = -18

⇒ y = $\dfrac{18}{10}$ = 1.8

AD = 5 - 1.8 = 3.2

Using equation (1),

⇒ 9 = (1.8)² + x²

⇒ x² = 9 - 3.24

⇒ x² = 5.76

⇒ x = $\sqrt{5.76}$

⇒ x = 2.4

(i) cos ∠DBC = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CD}{BD}\\[1em] = \dfrac{1.8}{2.4}\\[1em] = \dfrac{3}{4}$

Hence, tan ∠DBC = $\dfrac{3}{4}$.

(ii) sin ∠DBA = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AB}\\[1em] = \dfrac{3.2}{4}\\[1em] = \dfrac{4}{5}$

Hence, sin ∠DBA = $\dfrac{4}{5}$.

In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC.

Answer

In isosceles triangle ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\dfrac{18}{2}$ = 9

cos ∠ABC = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{BD}{AB}\\[1em] = \dfrac{9}{15}\\[1em] = \dfrac{3}{5}$

Hence, cos ∠ABC = $\dfrac{3}{5}$.

In the figure, given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find :

(i) sin B

(ii) tan C.

(iii) sin² B + cos² B

(iv) tan C - cot B

Answer

In isosceles Δ ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\dfrac{8}{2}$ = 4

In Δ ABD,

⇒ AB² = BD² + AD² (∵ AB is hypotenuse)

⇒ 5² = 4² + AD²

⇒ 25 = 16 + AD²

⇒ AD² = 25 - 16

⇒ AD² = 9

⇒ AD = $\sqrt{9}$

⇒ AD = 3

(i) sin B = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{AD}{AB} = \dfrac{3}{5}$

Hence, sin B = $\dfrac{3}{5}$

(ii) tan C = $\dfrac{Perpendicular}{Base}$

$\dfrac{AD}{DC} = \dfrac{3}{4}$

Hence, tan C = $\dfrac{3}{4}$

(iii) sin² B + cos² B

sin B = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{AD}{AB} = \dfrac{3}{5}$

cos B = $\dfrac{Base}{Hypotenuse}$

$\dfrac{BD}{AB} = \dfrac{4}{5}$

Now, sin² B + cos² B

$= \Big(\dfrac{3}{5}\Big)^2 + \Big(\dfrac{4}{5}\Big)^2\\[1em] = \dfrac{9}{25} + \dfrac{16}{25}\\[1em] = \dfrac{9 + 16}{25}\\[1em] = \dfrac{25}{25}\\[1em] = 1$

Hence, sin² B + cos² B = 1.

(iv) tan C - cot B

tan C = $\dfrac{Perpendicular}{Base}$

$= \dfrac{AD}{DC} = \dfrac{3}{4}$

cot B = $\dfrac{Base}{Perpendicular}$

$= \dfrac{BD}{AD} = \dfrac{4}{3}$

Now, tan C - cot B

$= \dfrac{3}{4} - \dfrac{4}{3}\\[1em] = \dfrac{3 \times 3}{4 \times 3}- \dfrac{4 \times 4}{3 \times 4}\\[1em] = \dfrac{9}{12} - \dfrac{16}{12}\\[1em] = \dfrac{9 - 16}{12}\\[1em] = \dfrac{-7}{12}$

Hence, tan C - cot B = $\dfrac{-7}{12}$.

In triangle ABC; ∠ABC = 90°, ∠CAB = x°, $\text{tan x°} = \dfrac{3}{4}$ and BC = 15 cm. Find the measures of AB and AC.

Answer

Given:

$\text{tan x°} = \dfrac{3}{4}\\[1em] \text{tan x°} = \dfrac{Perpendicular}{Base} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{BC}{AB} = \dfrac{3}{4}$

∴ If length of AB = 4x unit, length of BC = 3x unit.

BC = 15 cm (∵ Given)

∴ 3x = 15

⇒ x = $\dfrac{15}{3}$ = 5 cm

∴ AB = 4x = 4 x 5 = 20 cm

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AB is hypotenuse)

⇒ AC² = (15)² + (20)²

⇒ AC² = 225 + 400

⇒ AC² = 625

⇒ AC = $\sqrt{625}$

⇒ AC = 25 cm

Hence, AB = 20 cm and AC = 25 cm.

Using the measurements given in the following figure :

(i) Find the value of sin Φ and tan θ.

(ii) Write an expression for AD in terms of θ.

Answer

(i) In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ 13² = 12² + CD²

⇒ 169 = 144 + CD²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$

⇒ CD = 5

sin Φ = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{CD}{BD} = \dfrac{5}{13}$

Draw a line parallel to BC from point D such that it meets AB at point E. This line DE will be ⊥ to AB.

From figure,

DE = BC = 12

In Δ BED,

⇒ BD² = BE² + DE² (∵ BD is hypotenuse)

⇒ 13² = BE² + 12²

⇒ 169 = BE² + 144

⇒ BE² = 169 - 144

⇒ BE² = 25

⇒ BE = $\sqrt{25}$

⇒ BE = 5

And, AE = AB - BE = 14 - 5 = 9

tan θ = $\dfrac{Base}{Perpendicular}$

$= \dfrac{DE}{AE} = \dfrac{12}{9} = \dfrac{4}{3}$

Hence, sin Φ = $\dfrac{5}{13}$ and tan θ = $\dfrac{4}{3}$

(ii) sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

sin θ = $\dfrac{DE}{AD} = \dfrac{12}{AD}$

AD = $\dfrac{12}{\text{sin θ}}$ = 12 cosec θ

cos θ = $\dfrac{Base}{Hypotenuse}$

cos θ = $\dfrac{AE}{AD} = \dfrac{9}{AD}$

AD = $\dfrac{9}{\text{cos θ}}$ = 9 sec θ

Hence, AD = 12 cosec θ or 9 sec θ.

In the given figure; BC = 15 cm and $\text{sin B} = \dfrac{4}{5}$.

(i) Calculate the measures of AB and AC.

(ii) Now, if tan ∠ADC = 1; calculate the measures of CD and AD.

Answer

(i) Given:

$\text{sin B} = \dfrac{4}{5}\\[1em] \text{sin B} = \dfrac{Perpendicular}{Hypotenuse} = \dfrac{4}{5}$

∴ If length of AC = 4x cm, length of AB = 5x cm.

In Δ ABC,

⇒ AB²= BC² + AC² (∵ AB is hypotenuse)

⇒ (5x)² = BC² + (4x)²

⇒ 25x² = BC² + 16x²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = $\sqrt{9\text{x}^2}$

⇒ BC = 3x

It is given that BC = 15 cm

3x = 15

x = $\dfrac{15}{3}$

x = 5 cm

AB = 5x = 5 x 5 cm = 25 cm

AC = 4x = 4 x 5 cm = 20 cm

Hence, AB = 25 cm and AC = 20 cm.

(ii) tan ∠ADC = 1

$\text{tan ∠ADC} = \dfrac{Perpendicular}{Base} = 1$

∴ If length of AC = x unit, length of CD = x unit.

From (i), we know AC = 20 cm

∴ x = 20 cm

So, AC = CD = 20 cm

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ AD² = 20² + 20²

⇒ AD² = 400 + 400

⇒ AD² = 800

⇒ AD = $\sqrt{800}$

⇒ AD = $20\sqrt{2}$

Hence, CD = 20 cm and AD = 20 $\sqrt{2}$ cm.

If sin A + cosec A = 2; find the value of sin² A + cosec² A.

Answer

sin A + cosec A = 2

Squaring both sides,

(sin A + cosec A)² = 2²

⇒ sin² A + cosec² A + 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A + 2 x sin A x $\dfrac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A + 2 x ${\cancel{\text{sin A}}}$x $\dfrac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A + 2 = 4

⇒ sin² A + cosec² A = 4 - 2

⇒ sin² A + cosec² A = 2

Hence, sin² A + cosec² A = 2.

If tan A + cot A = 5; find the value of tan² A + cot² A.

Answer

tan A + cot A = 5

Squaring both sides,

(tan A + cot A)² = 5²

⇒ tan² A + cot² A + 2 x tan A x cot A = 25

⇒ tan² A + cot² A + 2 x tan A x $\dfrac{1}{\text{tan A}}$ = 25

⇒ tan² A + cot² A + 2 x ${\cancel{\text{tan A}}}$x $\dfrac{1}{\cancel{\text{tan A}}}$ = 25

⇒ tan² A + cot² A + 2 = 25

⇒ tan² A + cot² A = 25 - 2

⇒ tan² A + cot² A = 23

Hence, tan² A + cot² A = 23.

Given : 4 sin θ = 3 cos θ; find the value of :

(i) sin θ

(ii) cos θ

(iii) cot² θ - cosec² θ

(iv) 4 cos² θ - 3 sin² θ + 2

Answer

Given:

4 sin θ = 3 cos θ

$⇒ \dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}}\\[1em] ⇒ \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4}\\[1em]$

∴ If length of BC = 3x unit, length of AB = 4x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25 \text{x}^2}$

⇒ AC = 5x

(i) sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{CB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$

Hence, sin θ = $\dfrac{3}{5}$.

(ii) cos θ = $\dfrac{Base}{Hypotenuse}$

$\dfrac{AB}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

Hence, cos θ = $\dfrac{4}{5}$.

(iii) cot² θ - cosec² θ + 2

cot θ = $\dfrac{Base}{Perpendicular}$

$\dfrac{AB}{BC} = \dfrac{4x}{3x} = \dfrac{4}{3}$

cosec θ = $\dfrac{Hypotenuse}{Perpendicular}$

$\dfrac{AC}{CB} = \dfrac{5x}{3x} = \dfrac{5}{3}$

Now, cot² θ - cosec² θ

$= \Big(\dfrac{4}{3}\Big)^2 - \Big(\dfrac{5}{3}\Big)^2 \\[1em] = \dfrac{16}{9} - \dfrac{25}{9} \\[1em] = \dfrac{16 - 25}{9} \\[1em] = \dfrac{- 9}{9} \\[1em] = -1$

Hence, cot² θ - cosec² θ + 2 = -1.

(iv) 4 cos² θ - 3 sin² θ + 2

cos θ = $\dfrac{Base}{Hypotenuse}$

$\dfrac{AB}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{CB}{AC} = \dfrac{3x}{5x} = \dfrac{3}{5}$

Now, 4 cos² θ - 3 sin² θ + 2

$= 4 \times \Big(\dfrac{4}{5}\Big)^2 - 3 \times \Big(\dfrac{3}{5}\Big)^2 + 2\\[1em] = 4 \times \dfrac{16}{25} - 3 \times \dfrac{9}{25} + 2\\[1em] = \dfrac{64}{25} - \dfrac{27}{25} + 2\\[1em] = \dfrac{64 - 27}{25} + 2\\[1em] = \dfrac{37}{25} + 2\\[1em] = \dfrac{37}{25} + \dfrac{2 \times 25}{25}\\[1em] = \dfrac{37}{25} + \dfrac{50}{25}\\[1em] = \dfrac{37 + 50}{25}\\[1em] = \dfrac{87}{25}\\[1em] = 3\dfrac{12}{25}$

Hence, 4 cos² θ - 3 sin² θ + 2 =$3\dfrac{12}{25}$.

Given : 17 cos θ = 15; find the value of tan θ + 2 sec θ.

Answer

Given:

17 cos θ = 15

cos θ = $\dfrac{15}{17}$

$\Rightarrow \text{cos θ} = \dfrac{Base}{Hypotenuse} = \dfrac{15}{17}\\[1em]$

∴ If length of AC = 17x unit, length of AB = 15x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ (17x)² = BC² + (15x)²

⇒ 289x² = BC² + 225x²

⇒ BC² = 289x² - 225x²

⇒ BC = $\sqrt{64 \text{x}^2}$

⇒ BC = 8x

tan θ = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CB}{AB} = \dfrac{8x}{15x} = \dfrac{8}{15}$

sec θ = $\dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{17x}{15x} = \dfrac{17}{15}$

Now, tan θ + 2 sec θ

$= \dfrac{8}{15} + 2 \times \dfrac{17}{15}\\[1em] = \dfrac{8}{15} + \dfrac{34}{15}\\[1em] = \dfrac{8 + 34}{15}\\[1em] = \dfrac{42}{15}\\[1em] = \dfrac{14}{5}\\[1em] = 2\dfrac{4}{5}$

Hence, tan θ + 2 sec θ = $2\dfrac{4}{5}$.

Given: 5 cos A - 12 sin A = 0; evaluate :

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}$

Answer

Given:

5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

⇒ $\dfrac{\text{sin A}}{\text{cos A}} = \dfrac{5}{12}$

⇒ $\text{tan A} = \dfrac{5}{12}$

$⇒ \text{tan A} = \dfrac{Perpendicular}{Base} = \dfrac{5}{12} \\[1em]$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169 \text{x}^2}$

⇒ AC = 13x

sin A = $\dfrac{Perpendicular}{Hypotenuse}$

$\dfrac{CB}{AC} = \dfrac{5x}{13x} = \dfrac{5}{13}$

cos A = $\dfrac{Base}{Hypotenuse}$

$\dfrac{AB}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$

Now,

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}\\[1em] = \dfrac{\dfrac{5}{13} + \dfrac{12}{13}}{2 \times \dfrac{12}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{5 + 12}{13}}{\dfrac{24}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{24 - 5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{19}{13}}\\[1em] = \dfrac{\dfrac{17}{\cancel{13}}}{\dfrac{19}{\cancel{13}}}\\[1em] = \dfrac{17}{19}$

Hence, $\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}} = \dfrac{17}{19}$.