PSEB 12th Class Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

Direction (1 – 4): Prove the following.

Question 1.
3 sin^-1 x = sin^-1 (3x – 4x³), x ∈ [- $\frac{1}{2}$, $\frac{1}{2}$]
Solution.
Let x = sin θ. Then, sin^-1 x = θ.
We have,
R.H.S. = sin^-1 (3x – 4x³) = sin^-1(3 sin θ – 4 sin³ θ)
= sin^-1 (sin 3θ) = 3θ = 3 sin^-1 x
= L.H.S.
Hence proved.

Question 2.
3 cos^-1 x = cos^-1 (4x³ – 3x), x ∈ [$\frac{1}{2}$, 1]
Solution.
Let x = cos θ. Then, cos^-1 x = θ.
We have, R.H.S. = cos^-1 (4x³ – 3x)
= cos^-1 (4cos 3θ – 3 cos θ)
= cos^-1 (cos 3θ) = 3θ = 3 cos^-1 x
= L.H.S.
Hence proved.

Question 3.
tan^-1 $\frac{2}{11}$ + tan^-1 $\frac{7}{24}$ = tan^-1 $\frac{1}{2}$.
Solution.
Given, tan^-1 $\frac{2}{11}$ + tan^-1 $\frac{7}{24}$ = tan^-1 $\frac{1}{2}$

Question 4.
2 tan^-1 $\frac{1}{2}$ + tan^-1 $\frac{1}{7}$ = tan^-1 $\frac{31}{17}$
Solution.
Given, 2 tan^-1 $\frac{1}{2}$ + tan^-1 $\frac{1}{7}$ = tan^-1 $\frac{31}{17}$
L.H.S. = 2 tan^-1 $\frac{1}{2}$ + tan^-1 $\frac{1}{7}$
= $\tan ^{-1}\left[\frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right]+\tan ^{-1}\left(\frac{1}{7}\right)$ [∵ 2 tan^-1 x = tan^-1 ($\frac{2 x}{1-x^{2}}$)]

Direction (5 – 10):- Write the following functions in the simplest form:

Question 5.
tan^-1 $\frac{\sqrt{1+x^{2}}-1}{x}$, x ≠ 0.
Solution.
We have, tan^-1 $\frac{\sqrt{1+x^{2}}-1}{x}$
put x = tan θ
⇒ θ = tan^-1 x

Question 6.
tan^-1 $\frac{1}{\sqrt{x^{2}-1}}$, |x| > 1
Solution.
Let x = sec θ, then θ = sec^-1 x
∴ tan^-1 $\frac{1}{\sqrt{x^{2}-1}}$ = tan^-1 $\left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)$
= tan^-1 $\left(\frac{1}{\sqrt{\tan ^{2} \theta}}\right)$ [∵ sec² θ – 1 = tan² θ]
= tan^-1 $\left(\frac{1}{\tan \theta}\right)$
= tan^-1 (cot θ)
= tan^-1 [tan ($\frac{\pi}{2}$ – θ)] [∵ tan ($\frac{\pi}{2}$ – θ) = cot θ]
= $\frac{\pi}{2}$ – θ
= $\frac{\pi}{2}$ – sec^-1 x

Question 7.
tan^-1 $\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$, x < π.
Solution.
We have, tan^-1 $\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$
= tan^-1 $\left(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\right)$
= tan^-1 (tan $\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$)
= tan^-1 (tan $\frac{x}{2}$)
= $\frac{x}{2}$.

Question 8.
tan^-1 ($\frac{\cos x-\sin x}{\cos x+\sin x}$), 0 < x < π.
Solution.

Question 9.
tan^-1 $\frac{x}{\sqrt{a^{2}-x^{2}}}$, |x| < a.
Solution.
We have, tan^-1 $\frac{x}{\sqrt{a^{2}-x^{2}}}$
Let x = a sin θ
⇒ $\frac{x}{a}$ = sin θ
⇒ θ = sin^-1 ($\frac{x}{a}$)
∴ tan^-1 $\frac{x}{\sqrt{a^{2}-x^{2}}}$ = tan^-1 $\left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)$
= tan^-1 $\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right)$
= tan^-1 $\left(\frac{a \sin \theta}{a \cos \theta}\right)$
= tan^-1 (tan θ)
= θ = sin^-1 $\frac{x}{a}$.

Question 10.
tan^-1 $\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$, a > 0; $\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$.
Solution.
We have, tan^-1 $\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$, a > 0; $\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$
Let x = a tan θ
⇒ $\frac{x}{a}$ = tan θ
⇒ θ = tan^-1 $\frac{x}{a}$

∴ tan^-1 $\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$, a > 0; $\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$ = tan^-1 $\left(\frac{3 a^{2} \cdot(a \tan \theta)-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot\left(a^{2} \tan ^{2} \theta\right)}\right)$

= tan ^-1 $\left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right)$

= tan^-1 $\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$

= tan^-1 (tan 3θ) [∵ tan 3θ = $\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}$]

= 3θ = 3 tan^-1 $\frac{x}{a}$.

Direction (11 – 15) : Find the value of each of the following.

Question 11.
tan^-1 [2 cos(2 sin^-1 $\frac{1}{2}$)].
Solution.
Let sin^-1 $\frac{1}{2}$ = x
Then, sin x = $\frac{1}{2}$ = sin ($\frac{\pi}{6}$))
Now, tan^-1 [2 cos(2 sin^-1 $\frac{1}{2}$)] = tan^-1 [2 cos(2 × $\frac{\pi}{6}$)]
= tan^-1 [2 cos $\frac{\pi}{3}$]
= tan^-1 [2 × $\frac{1}{2}$] [∵ cos ($\frac{\pi}{3}$) = $\frac{1}{2}$)
= tan^-1 1 = $\frac{\pi}{4}$.

Question 12.
cot(tan^-1 a + cot^-1 a).
Solution.
We have, cot(tan^-1 a + cot^-1 a)
= cot ($\frac{\pi}{2}$)
= 0 [∵ tan^-1 x + cot^-1 x = $\frac{\pi}{2}$]

Question 13.
tan $\frac{1}{2}$ [sin^-1 $\frac{2 x}{1+x^{2}}$ + cos^-1 $\frac{1-y^{2}}{1+y^{2}}$], |x| < 1, y > 0 and xy < 1.
Solution.
Let x = tan θ.
Then, θ = tan^-1 x.
∴ sin^-1 $\frac{2 x}{1+x^{2}}$ = sin^-1 $\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
= sin^-1 (sin 2θ) = 2θ = 2 tan^-1 x
Again, let y = tan φ.
Then, φ = tan^-1 y
∴ cos^-1 $\frac{1-y^{2}}{1+y^{2}}$ = cos^-1 $\left(\frac{1-\tan ^{2} \varphi}{1+\tan ^{2} \varphi}\right)$
= cos^-1 (cos 2φ) = 2φ = 2 tan^-1 y
Now, tan $\frac{1}{2}$ [sin^-1 $\frac{2 x}{1+x^{2}}$ + cos^-1 $\frac{1-y^{2}}{1+y^{2}}$]
= tan $\frac{1}{2}$ [2 tan^-1 x + tan^-1 y]
= tan [tan^-1 x + tan^-1 y]
= tan[tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
[∵ tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= $\frac{x+y}{1-x y}$

Question 14.
If sin(sin^-1 $\frac{1}{5}$ + cos^-1 x) = 1, then find the value of x.
Solution.
Given, sin(sin^-1 $\frac{1}{5}$ + cos^-1 x) = 1
⇒ sin^-1 $\frac{1}{5}$ + cos^-1 x = sin^-1 (1)
[∵ sin θ = x ⇒ θ = sin^-1 x]
⇒ sin^-1 $\frac{1}{5}$ + cos^-1 x = sin^-1 (sin $\frac{\pi}{2}$)
[∵ sin ($\frac{\pi}{2}$) = 1]
⇒ sin^-1 $\frac{1}{5}$ + cos^-1 x = $\frac{\pi}{2}$
⇒ sin^-1 $\frac{1}{5}$ = $\frac{\pi}{2}$ – cos^-1 x
sin^-1 $\frac{1}{5}$ = sin^-1 x
[∵ sin^-1 x + cos^-1 x = $\frac{\pi}{2}$]
⇒ $\frac{1}{5}$ = x
Hence, the value of x is $\frac{1}{5}$.

Question 15.
If tan^-1 $\frac{x-1}{x-2}$ + tan^-1 $\frac{x+1}{x+2}=\frac{\pi}{4}$, then find the value of x.
Solution.
We have, tan^-1 $\frac{x-1}{x-2}$ + tan^-1 $\frac{x+1}{x+2}=\frac{\pi}{4}$

⇒

Direction (16 – 18): Find the value of each expression.

Question 16.
sin^-1 (sin $\frac{2 \pi}{3}$)
Solution.
We have, sin^-1 (sin $\frac{2 \pi}{3}$)
We know that sin^-1(sin x) = x , if x ∈ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$) which is the principal value branch of sin^-1 x.
Here, $\frac{2 \pi}{3}$ ∉ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$
Now, sin^-1 (sin $\frac{2 \pi}{3}$) can be written as
sin^-1 (sin $\frac{2 \pi}{3}$) = $\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$ where $\frac{\pi}{3}$ ∈ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$)
∴ sin^-1 (sin $\frac{2 \pi}{3}$) = sin^-1 (sin $\frac{\pi}{3}$) = $\frac{\pi}{3}$

Question 17.
tan^-1 (tan $\frac{3 \pi}{4}$)
Solution.
We have, tan^-1 (tan $\frac{3 \pi}{4}$)
We know that tan^-1 (tan x) = x, if x ∈ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$), which is the principal value branch of tan^-1 x.
Here, $\frac{3 \pi}{4}$ ∉ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$)
Now, tan^-1 (tan $\frac{3 \pi}{4}$) can be written as
tan^-1 (tan $\frac{3 \pi}{4}$) = tan^-1 [tan(π – $\frac{\pi}{4}$)]
= tan^-1 [- tan $\frac{\pi}{4}$]
= tan^-1 [tan (- $\frac{\pi}{4}$)]
where – $\frac{\pi}{4}$ ∈ (- $\frac{\pi}{2}$, $\frac{\pi}{2}$)
[∵ – tan θ = tan(- θ)]
∴ tan^-1 (tan $\frac{3 \pi}{4}$) = tan^-1 [tan (- $\frac{\pi}{4}$)]
= – $\frac{\pi}{4}$.

Question 18.
tan (sin^-1 $\frac{3}{5}$ + cot^-1 $\frac{3}{2}$)
Solution.
Let sin^-1 $\frac{3}{5}$ = x.
Then, sin x = $\frac{3}{5}$
⇒ cos x = $\sqrt{1-\sin ^{2} x}$ = $\frac{4}{5}$
⇒ sec x = $\frac{5}{4}$
∴ tan x = $\sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}$
∴ x = tan^-1 $\frac{3}{4}$
∴ sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{3}{4}$ ………..(i)
Now, cot^-1 $\frac{3}{2}$ = tan^-1 $\frac{2}{3}$
[∵ tan^-1 $\frac{1}{x}$ = cot^-1 x] ……………(ii)
Hence, tan (sin^-1 $\frac{3}{5}$ + cot^-1 $\frac{3}{2}$)
= tan (tan^-1 $\frac{3}{4}$ + tan^-1 $\frac{2}{3}$)
= $\tan \left(\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)$
[∵ tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= tan (tan^-1 $\frac{9+8}{12-6}$)
= tan (tan^-1 $\frac{17}{6}$)
= $\frac{17}{6}$.

Question 19.
cos^-1 (cos $\frac{7 \pi}{6}$) is equal to
(A) $\frac{7 \pi}{6}$

(B) $\frac{5 \pi}{6}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{6}$

Solution.
We know that cos^-1 (cos x) = x if x ∈ [0, x], which is the principal value branch of cos^-1 x.
Here, $\frac{7 \pi}{6}$ ∉ x ∈ [0, π]
Now, cos^-1 (cos $\frac{7 \pi}{6}$) can be written as
cos^-1 (cos $\frac{7 \pi}{6}$) = cos^-1 [cos(2π – $\frac{5 \pi}{6}$)]

= cos^-1 [cos $\frac{5 \pi}{6}$], where $\frac{5 \pi}{6}$ ∈ [0, π]
[∵ cos(2π – x) = cos x]
∴ cos^-1 (cos $\frac{7 \pi}{6}$) = cos^-1 (cos $\frac{5 \pi}{6}$)
= $\frac{5 \pi}{6}$
The correct option is (B).

Question 20.
sin[$\frac{\pi}{3}$ – sin^-1 (- $\frac{1}{2}$)] is equal to
(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $\frac{1}{4}$

(D) 1
Solution.
Let sin^-1 (- $\frac{1}{2}$) = x.
Then, sin x = – $\frac{1}{2}$ = – sin $\frac{\pi}{6}$ = sin(-$\frac{\pi}{6}$)
We know that the range of the principal value of sin^-1 x is (- $\frac{\pi}{2}$, –$\frac{\pi}{2}$)
∴ sin^-1 (- $\frac{1}{2}$) = – $\frac{\pi}{6}$
Now, sin[$\frac{\pi}{6}$ – sin^-1 (- $\frac{1}{2}$)] = sin $\left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]$
= sin $\left(\frac{\pi}{3}+\frac{\pi}{6}\right)$
= sin ($\frac{3 \pi}{6}$)
= sin ($\frac{\pi}{2}$) = 1
The correct option is (D).

Question 21.
tan^-1 (- √3) – cot^-1 (- √3) is equal to
(A) π
(B) – $\frac{\pi}{2}$
(C) 0
(D) 2√3
Solution.
Let tan^-1 √3 = x
⇒ tan x = √3 = tan $\frac{\pi}{3}$
∴ tan^-1 √3 = $\frac{\pi}{3}$
Again, let cos^-1(- √3) = x
⇒ cot x = – √3 = – cot $\frac{\pi}{6}$
= cot (π – $\frac{\pi}{6}$)
= cot $\frac{5 \pi}{6}$
∴ cot^-1 (- √3) = $\frac{5 \pi}{6}$
Now, tan^-1 (- √3) – cot^-1 (- √3) = $\frac{\pi}{3}$ – $\frac{5 \pi}{6}$
= $\frac{2 \pi-5 \pi}{6}=\frac{-3 \pi}{6}=-\frac{\pi}{2}$
Hence, correct option is (B).