ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions Chapter Test

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions Chapter Test

Question 1.
Show that sin^-1 \(\frac{\sqrt{3}}{2}\) + 2 tan^-1 \(\frac{1}{\sqrt{3}}\) = \(\frac{2 \pi}{3}\)
Solution:
Let sin^-1 \(\frac{\sqrt{3}}{2}\) = y ; y ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
sin y = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ sin^-1 \(\frac{\sqrt{3}}{2}\) = \(\frac{\pi}{3}\) …………..(1)
Now tan^-1 \(\left(\frac{1}{\sqrt{3}}\right)\) = tan^-1 (tan \(\frac{\pi}{6}\))
= \(\frac{\pi}{6}\) ……………..(2)
[∵ tan^-1 (tan x) = x ∀ x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]
Thus, sin^-1 \(\frac{\sqrt{3}}{2}\) + 2 tan^-1 \(\frac{1}{\sqrt{3}}\)
= \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\)
= \(\frac{2 \pi}{3}\) [Using (1) and (2)]

Question 2.
Find the value of the following:
(i) tan (cos^-1 (cos \(\frac{7 \pi}{6}\)))
(ii) cosec (cos^-1 (- \(\frac{4}{5}\)))
Solution:
(i) tan (cos^-1 (cos \(\frac{7 \pi}{6}\)))
= tan {cos^-1 (cos (2π – \(\frac{5 \pi}{6}\)))}
= tan {cos^-1 (cos \(\frac{5 \pi}{6}\))}
[∵ cos^-1 (cos θ) = θ ∀ θ ∈ [0, π]]
= tan \(\frac{5 \pi}{6}\)
= tan (π – \(\frac{\pi}{6}\))
= – tan \(\frac{\pi}{6}\)
= – \(\frac{1}{\sqrt{3}}\)

(ii) cosec {cos^-1 (- \(\frac{4}{5}\))}
= cosec {π – cos^-1 \(\frac{4}{5}\)}
[∵ cos^-1 (- x) = π – cos^-1 x]
put cos^-1 \(\frac{4}{5}\) = θ
⇒ cos θ = \(\frac{4}{5}\)
= cosec ( – θ) = + cosec θ
= + \(\frac{1}{\sin \theta}\)
= + \(\frac{i}{\sqrt{1-\cos ^2 \theta}}\)
= + \(\frac{1}{\sqrt{1-\left(\frac{4}{5}\right)^2}}\)
= + \(\frac{1}{\sqrt{\frac{25-i 6}{25}}}\)
= + \(\frac{5}{3}\)

Question 3.
Find the value of cos (2 cos^-1 x + sin^-1 x) at x = \(\frac{1}{5}\).
Solution:
cos (2 cos^-1 x + sin^-1 x) = cos (cos^-1 x + cos^-1 x + sin^-1 x)
= cos (cos^-1 x + \(\frac{\pi}{2}\))
[∵ cos^-1 x + sin^-1 x = \(\frac{\pi}{2}\) ∀ x ∈ [- 1, 1]]

= cos (\(\frac{\pi}{2}\) + cos^-1 x)
= – sin (cos^-1 x)
= – \(\sqrt{1-x^2}\)
at x = \(\frac{1}{5}\) ;
given expression = – \(\sqrt{1-\left(\frac{1}{5}\right)^2}\)
= \(-\sqrt{1-\frac{1}{25}}=-\sqrt{\frac{24}{25}}\)
= – \(\frac{2}{5}\) √6

Question 4.
Prove that cos^-1 x = 2 sin^-1 \(\sqrt{\frac{1-x}{2}}\) = 2 cos^-1 \(\sqrt{\frac{1+x}{2}}\).
Solution:
Putting x = cos θ
⇒ θ = cos^-1 x where θ ∈ [0, π]

Question 5.
Prove that :
(i) tan^-1 \(\frac{2}{3}\) = \(\frac{1}{2}\) tan^-1 \(\frac{12}{5}\)
(ii) tan^-1 \(\left(\frac{1}{2}\right)\) + tan^-1 \(\left(\frac{1}{5}\right)\) + tan^-1 \(\left(\frac{1}{8}\right)\) = \(\frac{\pi}{4}\)
(iii) tan^-1 \(\frac{1}{4}\) + tan^-1 \(\frac{2}{9}\) = sin^-1 \(\frac{1}{\sqrt{5}}\)
(iv) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{239}\) = \(\frac{\pi}{4}\)
Solution:
(i) L.H.S. = tan^-1 \(\frac{2}{3}\) = \(\frac{1}{2}\) tan^-1 \(\frac{12}{5}\)
= \(\frac{1}{2}\) tan^-1 \(\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^2}\right)\)
[∵ 2 tan^-1 x = tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\) if |x| < 1]
= \(\frac{1}{2}\) tan^-1 \(\left(\frac{4 / 3}{5 / 9}\right)\)
= \(\frac{1}{2}\) tan^-1 \(\left(\frac{12}{5}\right)\)
= R.H.S

(ii) L.H.S. = tan^-1 \(\left(\frac{1}{2}\right)\) + tan^-1 \(\left(\frac{1}{5}\right)\) + tan^-1 \(\left(\frac{1}{8}\right)\)

(iii) L.H.S. = tan^-1 \(\frac{1}{4}\) + tan^-1 \(\frac{2}{9}\)
= tan^-1 \(\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \cdot \frac{2}{9}}\right)\)
[∵ tan^-1x – tan^-1 y = tan^-1 \(\left(\frac{x-y}{1+x y}\right)\) if xy < 1]

(iv) L.H.S. = 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{239}\)

Question 6.
Prove the following:
(i) 2 sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{24}{7}\) (NCERT)
(ii) sin^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{2}{\sqrt{5}}\) = cot^-1 \(\frac{2}{11}\)
(iii) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
(iv) sin^-1 \(\frac{8}{17}\) + sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{77}{36}\) (NCERT)
(v) sin^-1 \(\frac{8}{17}\) + cos^-1 \(\frac{4}{5}\) = cot^-1 \(\frac{36}{77}\)
Solution:
(i) L.H.S. = 2 sin^-1 \(\frac{3}{5}\)
= sin^-1 \(\left[2 \times \frac{3}{5} \times \sqrt{1-\frac{9}{25}}\right]\)

[∵ 2 sin^-1 x = sin^-1 (2x \(\sqrt{1-x^2}\))
if \(-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)]

= sin^-1 \(\left[2 \times \frac{3}{5} \times \frac{4}{5}\right]\)
= sin^-1 \(\left(\frac{24}{25}\right)\)
= tan^-1 \(\left(\frac{24}{7}\right)\)
= R.H.S.

(ii) First of all, we convert sin to tan, we make the right angled Δ with

p = 4 ; h = 5 ; b = 3
∴ sin^-1 \(\frac{4}{5}\) = tan^-1 \(\left(\frac{4}{3}\right)\) …………(1)
Also convert cos^-1 to tan^-1, we make the right angled Δ with
b = 2 ; h = √5
∴ p = \(\sqrt{5-4}\) = 1
∴ cos^-1 \(\left(\frac{2}{\sqrt{5}}\right)\) = tan^-1 \(\left(\frac{1}{2}\right)\) …………(2)

L.H.S. = sin^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{2}{\sqrt{5}}\)
= tan^-1 \(\frac{4}{3}\) + tan^-1 \(\frac{1}{2}\) [Using (1) and (2)]
= tan^-1 \(\left[\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3} \cdot \frac{1}{2}}\right]\)
[∵ tan^-1x – tan^-1 y = tan^-1 \(\left(\frac{x-y}{1+x y}\right)\) if xy > – 1
Here, xy = \(\frac{4}{3} \cdot \frac{1}{2}=\frac{2}{3}\) < 1]
= tan^-1 \(\left[\frac{\frac{11}{6}}{\frac{2}{6}}\right]\)
= tan^-1 \(\left(\frac{11}{2}\right)\)
= cot^-1 \(\left(\frac{2}{11}\right)\)
[∵ tan^-1 \(\frac{1}{x}\) = cot^-1 x if x > 0]

(iii) Let sin^-1 \(\frac{3}{5}\) = θ
⇒ sin θ = \(\frac{3}{5}\)
⇒ cos θ = \(\sqrt{1-\frac{9}{25}}=\frac{4}{5}\) and
sin^-1 \(\frac{8}{17}\) = Φ
⇒ sin Φ = \(\frac{8}{17}\)
⇒ cos Φ = \(\sqrt{1-\frac{64}{269}}=\frac{15}{17}\)
Now cos (θ + Φ) = cos θ sin Φ – sin θ sin Φ
= \(\frac{4}{5} \times \frac{15}{17}-\frac{3}{5} \times \frac{8}{17}=\frac{36}{85}\)
⇒ θ + Φ = cos^-1 \(\)
⇒ sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\left(\frac{36}{85}\right)\)

(iv) put sin^-1 \(\frac{8}{17}\) = θ
⇒ \(\frac{8}{17}\) = sin θ
∴ cos θ = \(\sqrt{1-\frac{64}{289}}=\frac{15}{17}\)
put sin^-1 \(\frac{3}{5}\) = Φ
⇒ \(\frac{3}{5}\) = sin Φ
∴ cos Φ = \(\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
∴ tan θ = \(\frac{\sin \theta}{\cos \theta}\)
= \(\frac{\frac{8}{17}}{\frac{15}{17}}=\frac{8}{15}\) and
tan Φ = \(\frac{\sin \phi}{\cos \phi}\)
= \(\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\)
Now, tan (θ + Φ) = \(\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\)
= \(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\)
= \(\frac{\frac{77}{60}}{\frac{36}{60}}=\frac{77}{36}\)
∴ θ + Φ = tan^-1 \(\left(\frac{77}{36}\right)\)
⇒ sin^-1 \(\frac{8}{17}\) + sin^-1 \(\frac{3}{5}\) = tan^-1 \(\left(\frac{77}{36}\right)\)

(v) L.H.S. = sin^-1 \(\frac{8}{17}\) + cos^-1 \(\frac{4}{5}\)
= sin^-1 \(\frac{8}{17}\) + sin^-1 \(\frac{3}{5}\)

= sin^-1 \(\left[\frac{8}{17} \sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5} \sqrt{1-\left(\frac{8}{17}\right)^1}\right]\)
[∵ sin^-1 x + sin^-1 y = sin^-1 \(\left[x \sqrt{1-y^2}+y \sqrt{1-x^2}\right]\)]

= sin^-1 \(\left[\frac{8}{17} \times \frac{4}{5}+\frac{3}{5} \times \frac{15}{17}\right]\)
= sin^-1 \(\left(\frac{77}{85}\right)\)
= cot^-1 \(\left(\frac{36}{77}\right)\)
= R.H.S.

(v) sin^-1 \(\frac{x}{\sqrt{1-x^2}}\) + cos \(\frac{x+1}{\sqrt{x^2+2 x+2}}\) = tan^-1 (x² + x + 1). (old)
Solution:
First of all convert sin^-1 to tan^-1
we make the right angled triangle with p =- x ; h = \([\sqrt{1+x^2}/latex]

∴ b = [latex]\sqrt{h^2-p^2}\)
= \(\sqrt{1+x^2-x^2}\) = 1
∴ sin \(\left(\frac{x}{\sqrt{1+x^2}}\right)\) = tan x
Now we convert cos^-1 to tan^-1, for this we make the right angled Δ with

b = x + 1 ;
h = \(\sqrt{x^2+2 x-2}\)
p = \(\sqrt{h^2-b^2}\)
= \(\sqrt{x^2+2 x+2-x^2-2 x-1}\) = 1
Thus cos^-1 \(\left(\frac{x+1}{\sqrt{x^2+2 x+2}}\right)\) = tan^-1 \(\left(\frac{1}{x+1}\right)\) …………. (2)
∴ L.H.S. = sin^-1 \(\left(\frac{x}{\sqrt{1+x^2}}\right)\) + cos^-1 \(\left(\frac{x+1}{\sqrt{x^2+2 x+2}}\right)\)
= tan^-1 x + tan^-1 \(\left(\frac{1}{x+1}\right)\) [using (1) and (2)]
= tan^-1 \(\) [Here x . \(\frac{1}{x+1}\) < 1]
[∵ tan^-1x – tan^-1 y = tan^-1 \(\left(\frac{x-y}{1+x y}\right)\) if xy > – 1]
= tan^-1 \(\left[\frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}\right]\)
= tan^-1 (x² + x + 1)
= R.H.S.

Question 7.
Find the value of cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\).
Solution:
cot^-1 9 = tan^-1 \(\frac{1}{9}\)
[∵ cot^-1 x = tan^-1 \(\frac{1}{x}\) if x > 0]
We convert cosec^-1 to tan^-1, for this we make a right angled Δ with
h = \(\sqrt{41}\) ; p = 4

∴ b = \(\sqrt{h^2-p^2}\)
= \(\sqrt{41-16}\) = 5
∴ cosec^-1 \(\frac{\sqrt{41}}{4}\) = tan^-1 \(\left(\frac{4}{5}\right)\) ………(2)
Thus, cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = tan^-1 \(\frac{1}{9}\) + tan^-1 \(\frac{4}{5}\)
= tan^-1 \(\left[\frac{\frac{1}{9}+\frac{4}{5}}{1-\frac{1}{9} \cdot \frac{4}{5}}\right]\)
[∵ tan^-1x – tan^-1 y = tan^-1 \(\left(\frac{x-y}{1+x y}\right)\) if xy < 1 Here, xy = \(\frac{1}{9} \times \frac{4}{5}\) < 1]
= tan^-1 \(\left[\frac{\frac{5+36}{45}}{\frac{45-4}{45}}\right]\)
= tan^-1 \(\left(\frac{\frac{41}{45}}{\frac{41}{45}}\right)\)
= tan^-1 (1)
= tan^-1 (tan \(\frac{\pi}{4}\))
= \(\frac{\pi}{4}\)
[∵ tan^-1 (tan x) = x ∀ x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]

Question 8.
If two angles of a triangle are tan^-1 2 and tan^-1 3, find the third angle.
Solution:
Since the sum of the angle of a triangle be 180° i.e., π
∴ π – (tan^-1 2 + tan^-1 3)
= π – [π + tan^-1 \(\left(\frac{2+3}{1-2 \times 3}\right)\)]
[∵ tan^-1 x – tan^-1 y = tan^-1 \(\left(\frac{x-y}{1+x y}\right)\) if xy > – 1 Here xy = 2 × 3 = 6 > 1]
= π – π – tan^-1 \(\left(\frac{5}{-5}\right)\)
= – tan^-1 (- 1) = + tan^-1 (1)
[∵ tan^-1 (- x) = – tan^-1 x]
= tan^-1 (tan \(\frac{\pi}{4}\))
= \(\frac{\pi}{4}\)
[∵ tan^-1 (tan x) = x ∀ x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]