PSEB 12th Class Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

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

PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Find the value of the following :

Question 1.
cos^-1 (cos $\frac{13 \pi}{6}$).
Solution.
We know that cos^-1(cos x) = x if x ∈ [0, π], which is the principal value of cos^-1 x.
Here, $\frac{13 \pi}{6}$ ∉ [0, π].
Now, cos^-1 (cos $\frac{13 \pi}{6}$) can be written as
cos^-1 (cos $\frac{13 \pi}{6}$) = cos^-1 [cos (2π + $\frac{pi}{6}$)]
= cos^-1 [cos ($\frac{pi}{6}$)],
where $\frac{pi}{6}$ ∈ [0, π]
[∵ cos(2π + x) = cos x]
∴ cos^-1 (cos $\frac{13 \pi}{6}$) = cos^-1 [cos ($\frac{pi}{6}$)]
= $\frac{pi}{6}$.

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

Prove that

Question 3.
2 sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{24}{7}$.
Solution.
Let sin^-1 $\frac{3}{5}$ = x.
Then, sin x = $\frac{3}{5}$
⇒ cos x = $\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
∴ tan x = $\frac{3 / 5}{4 / 5}=\frac{3}{4}$
∴ x = tan^-1 $\frac{3}{4}$
⇒ sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{3}{4}$
Now, we have
L.H.S = 2 sin^-1 $\frac{3}{5}$ = 2 tan^-1 $\frac{3}{4}$
= tan^-1 $\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)$

[∵ 2 tan^-1 x = tan^-1 $\frac{2 x}{1-x^{2}}$]

= tan^-1 $\left(\frac{\frac{3}{2}}{\frac{16-9}{16}}\right)$

= tan^-1 $\left(\frac{3}{2} \times \frac{16}{7}\right)$

= tan^-1 $\frac{24}{7}$

= R.H.S.
Hence proved.

Question 4.
sin^-1 $\frac{8}{17}$ + sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{77}{36}$
solution.
Let sin^-1 $\frac{8}{17}$ = x.
Then, sin x = $\frac{8}{17}$
⇒ cos x = $\sqrt{1-\left(\frac{8}{17}\right)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$
∴ tan x = $\frac{8 / 17}{15 / 17}=\frac{8}{15}$
⇒ x = tan^-1 $\frac{8}{15}$
∴ sin^-1 $\frac{8}{17}$ = tan^-1 $\frac{8}{15}$ …………..(i)

Now, let sin^-1 $\frac{3}{5}$ = y.
Then, sin y = $\frac{3}{5}$
⇒ cos y = $\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
∴ tan y = $\frac{3 / 5}{4 / 5}=\frac{3}{4}$
⇒ y = tan^-1 $\frac{3}{4}$
∴ sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{3}{4}$ …………..(ii)

Now, we have
L.H.S = sin^-1 $\frac{8}{17}$ + sin^-1 $\frac{3}{5}$
[Using Eqs. (i) and (ii)]
= tan^-1 $\frac{8}{15}$ + tan^-1 $\frac{3}{4}$
= tan^-1 $\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}$
= tan^-1 $\left(\frac{32+45}{60-24}\right)$
[tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= tan^-1 $\frac{77}{36}$
= R.H.S
Hence proved.

Question 5.
cos^-1 $\frac{4}{5}$ + cos^-1 $\frac{12}{13}$ = cos^-1 $\frac{33}{65}$
Solution.
Let cos^-1 $\frac{4}{5}$ = x
Then, cos x = $\frac{4}{5}$
⇒ sin x = $\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
∴ tan x = $\frac{3 / 5}{4 / 5}=\frac{3}{4}$ …………(i)

⇒ x = tan^-1 $\frac{3}{4}$
Now, let cos^-1 $\frac{12}{13}$ = y.
Then cos y = $\frac{12}{13}$
⇒ sin y = $\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
∴ tan y = $\frac{5 / 13}{12 / 13}=\frac{5}{12}$
⇒ y = tan^-1 $\frac{5}{12}$
∴ cos^-1 $\frac{12}{13}$ = tan^-1 $\frac{5}{12}$ ……………(ii)

Let cos^-1 $\frac{33}{65}$ = z.
Then, cos z = $\frac{33}{65}$
⇒ sin z = $\sqrt{1-\left(\frac{33}{65}\right)^{2}}=\sqrt{\frac{3136}{4225}}=\frac{56}{65}$
∴ tan z = $\frac{56 / 65}{33 / 65}=\frac{56}{33}$
⇒ z = tan^-1 $\frac{56}{33}$
∴ cos^-1 $\frac{33}{65}$ = tan^-1 $\frac{56}{33}$ …………..(iii)

Now, we have
L.H.S = cos^-1 $\frac{4}{5}$ + cos^-1 $\frac{12}{13}$
= $\frac{3 / 5}{4 / 5}=\frac{3}{4}$ + tan^-1 $\frac{5}{12}$
[∵ Usin Eqs. (i) and (ii)]
= tan^-1 $\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}$
[∵ tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= tan^-1 $\frac{36+20}{48-15}$
= tan^-1 $\frac{56}{33}$
= cos^-1 $\frac{33}{65}$ = R.H.S.
Hence proved.

Question 6.
cos^-1 $\frac{12}{13}$ + sin^-1 $\frac{3}{5}$ = sin^-1 $\frac{56}{65}$
Solution.
Let sin^-1 $\frac{3}{5}$ = x.
Then, sin x = $\frac{3}{5}$
⇒ cos x= $\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
∴ tan x = $\frac{3 / 5}{4 / 5}=\frac{3}{4}$
⇒ x = tan^-1 $\frac{3}{4}$
∴ sin^-1 $\frac{3}{5}$ = tan^-1 $\frac{3}{4}$ …………(i)

Now, let cos^-1 $\frac{12}{13}$ = y.
Then, cos y = $\frac{12}{13}$
⇒ sin y = $\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
∴ tan y = $\frac{5 / 13}{12 / 13}=\frac{5}{12}$
⇒ y = tan^-1 $\frac{5}{12}$
∴ cos^-1 $\frac{12}{13}$ = tan^-1 $\frac{5}{12}$ ………………(ii)

Let sin^-1 $\frac{56}{65}$ = z.
Then, sin z = $\frac{56}{65}$
⇒ cos z = $\sqrt{1-\left(\frac{56}{65}\right)^{2}}=\sqrt{\frac{1089}{4225}}=\frac{33}{65}$
∴ tan z = $\frac{56 / 65}{33 / 65}=\frac{56}{33}$
⇒ z = tan^-1 $\frac{56}{33}$
∴ sin^-1 $\frac{56}{65}$ = tan^-1 $\frac{56}{33}$ …………….(iii)

Now, we have
L.H.S. = cos^-1 $\frac{12}{13}$ + sin^-1 $\frac{3}{5}$
= tan^-1 $\frac{3}{4}$ + tan^-1 $\frac{5}{12}$
[Using Eqs. (i) and (ii)]
= tan^-1 $\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}}$

[∵ tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= tan^-1 $\frac{20+36}{48-15}$
= tan^-1 $\frac{56}{33}$
= sin^-1 $\frac{56}{65}$.
Hence proved.

Question 7.
tan^-1 $\frac{63}{16}$ = sin^-1 $\frac{5}{13}$ + cos ^-1 $\frac{3}{5}$
Solution.
Let sin^-1 $\frac{5}{13}$ = x
Then, sin x = $\frac{5}{13}$
⇒ cos x = $\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$
∴ tan x = $\frac{5 / 13}{12 / 13}=\frac{5}{12}$
⇒ x = tan^-1 $\frac{5}{12}$
∴ sin^-1 $\frac{5}{13}$ = tan^-1 $\frac{5}{12}$ ………….(i)

Let cos^-1 $\frac{3}{5}$ = y.
Then, cos y = $\frac{3}{5}$
⇒ sin y = $\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
∴ tan y = $\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
⇒ y = tan^-1 $\frac{4}{5}$
∴ cos^-1 $\frac{3}{5}$ = tan^-1 $\frac{4}{5}$ ………………(ii)

Using Eqs. (i) and (ii), we have
R.H.S = sin^-1 $\frac{5}{13}$ + cos^-1 $\frac{3}{5}$
= tan^-1 $\frac{5}{12}$ + tan^-1 $\frac{4}{5}$
[∵ tan^-1 x + tan^-1 y = tan^-1 $\left(\frac{x+y}{1-x y}\right)$]
= tan^-1 $\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right)$

= tan^-1 $\left(\frac{15+48}{36-20}\right)$

= tan^-1 $\frac{63}{16}$ = L.H.S.
Hence proved.

Question 8.
tan^-1 $\frac{1}{5}$ + tan^-1 $\frac{1}{7}$ + tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{8}$ = $\frac{\pi}{4}$
Solution.
tan^-1 $\frac{1}{5}$ + tan^-1 $\frac{1}{7}$ + tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{8}$

Question 9.
tan^-1 √x = $\frac{1}{2}$ cos^-1 $\left(\frac{1-x}{1+x}\right)$, x ∈ [0, 1]
Solution.
Let x = tan² θ.
Then, √x = tan θ
⇒ θ = tan^-1 √x
∴ $\frac{1-x}{1+x}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ = cos 2θ
[∵ cos 2θ = $\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$]
Now, we have
R.H.S = $\frac{1}{2}$ cos^-1 $\left(\frac{1-x}{1+x}\right)$
= $\frac{1}{2}$ cos^-1 2θ = θ
= tan^-1 √x = L.H.S
Hence Proved.

Question 10.
cot^-1 $\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$, x ∈ (0, $\frac{\pi}{4}$)
Solution.
Consider, $\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$

= $\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}}$ (By rationalising)

= $\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}$

= $\frac{2\left(1+\sqrt{\left.1-\sin ^{2} x\right)}\right.}{2 \sin x}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$

= cot $\frac{x}{2}$

L.H.S = cot^-1 $\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$

= cot^-1 (cot $\frac{x}{2}$)
= $\frac{x}{2}$ = R.H.S

Question 11.
tan^-1 $\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ = $\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$, $-\frac{1}{\sqrt{2}}$ ≤ x ≤ 1.
[Hint: put x = cos 2θ]
Solution.
Put x = cos 2θ, so that θ = $\frac{1}{2}$ cos^-1 x.
Then, we have

Question 12.
$\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$
Solution.
L.H.S = $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}$

= $\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)$

= $=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)$ [∵ sin^-1 x + cos^-1 x = $\frac{\pi}{2}$]

= $\frac{9}{4}\left(\sin ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^{2}}\right)$
[∵ cos^-1 x = sin^-1 $\sqrt{1-x^{2}}$]

= $\frac{9}{4} \sin ^{-1} \sqrt{\frac{8}{9}}$

= $\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

= R.H.S
Hence proved.

Direction (13 – 17): Solve the following equations.

Question 13.
2 tan^-1 (cos x) = tan^-1 (2cosec x)
Solution.
We have, 2 tan^-1 (cos x) = tan^-1 (2 cosec x)
⇒ tan^-1 $\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)$ = tan^-1 (2 cosec x)

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

⇒ $\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)$ = 2 cosec x

⇒ $\frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}$

⇒cos x = sin x

⇒ tan x = 1 = tan $\frac{\pi}{4}$.

Question 14.
tan^-1 $\frac{1-x}{1+x}$ = $\frac{1}{2}$ tan^-1 x, (x > 0)
Solution.
We have, sin^-1 (1 – x) – 2 sin^-1 x = $\frac{\pi}{2}$
⇒ – 2 sin^-1 x = $\frac{\pi}{2}$ – sin^-1 (1 – x)
⇒ – 2 sin^-1 x = cos^-1 (1 – x) ………….(i)
Let sin^-1 x = θ
⇒ sin θ = x
⇒ cos θ = $\sqrt{1-x^{2}}$
∴ θ = cos^-1 $\sqrt{1-x^{2}}$
∴ sin^-1 x = cos^-1 $\sqrt{1-x^{2}}$
Therefore, from Eq. (i), we have
– 2 cos^-1 ($\sqrt{1-x^{2}}$ ) = cos^-1 (1 – x)
Put x = sin y. Then, we have
– 2 cos^-1 () = cos^-1 (1 – sin y)
⇒ – 2 cos^-1 (cos y) = cos^-1 (1 – sin y)
⇒ – 2y = cos^-1 (1 – sin y)
⇒ 1 – sin y = cos(- 2y) = cos 2y
⇒ 1 – sin y = 1 – 2 sin2 y
⇒ 2 sin² y – sin y = 0
⇒ sin y(2 sin y – 1) = 0
sin y = 0 or $\frac{1}{2}$
∴ x = 0 or x = $\frac{1}{2}$
But, when x = $\frac{1}{2}$, it can be observed that
We have, tan^-1 $\frac{1-x}{1+x}$ = $\frac{1}{2}$ tan^-1 x
⇒ tan^-1 1 – tan^-1 x = $\frac{1}{2}$ tan^-1 x
[∵ tan^-1 x – tan^-1 y = tan^-1 $\frac{(x-y)}{1+x y}$]
[∵ tan^-1 (1) = $\frac{\pi}{4}$]
⇒ $\frac{\pi}{4}$ = $\frac{3}{2}$ tan^-1 x
⇒ tan^-1 x = $\frac{\pi}{6}$
⇒ x = tan $\frac{\pi}{6}$
∴ x = $\frac{1}{\sqrt{3}}$.

Question 15.
sin(tan^-1 x), |x| < 1 is equal to
(A) $\frac{x}{\sqrt{1-x^{2}}}$

(B) $\frac{1}{\sqrt{1-x^{2}}}$

(C) $\frac{1}{\sqrt{1+x^{2}}}$

(D) $\frac{x}{\sqrt{1+x^{2}}}$
Solution.
Let tan^-1 x = y.
Then, tan y = x
⇒ sin y = $\frac{x}{\sqrt{1+x^{2}}}$
∴ y = sin^-1 ($\frac{x}{\sqrt{1+x^{2}}}$)
⇒ tan^-1 x = sin^-1 ($\frac{x}{\sqrt{1+x^{2}}}$)
Now, sin(tan^-1 x) = sin(sin^-1 ($\frac{x}{\sqrt{1+x^{2}}}$))
= $\frac{x}{\sqrt{1+x^{2}}}$
The correct answer is (D).

Question 16.
sin^-1 (1 – x) – 2 sin^-1 x = $\frac{\pi}{2}$, then x is equal to
(A) 0, $\frac{1}{2}$
(B) 1, $\frac{1}{2}$
(C) 0
(D) $\frac{1}{2}$
Solution.
Given, sin^-1 (1 – x) – 2 sin^-1 x = $\frac{\pi}{2}$
putting $\frac{\pi}{2}$ = sin^-1 (1 – x) + cos^-1 (1 – x)
or sin^-1 (1 – x) – 2 sin^-1 (1 – x) = sin^-1 (1 – x) + cos^-1 (1 – x)
⇒ – 2 sin^-1 x = cos^-1 (1 – x)
Let sin^-1 x = α
∴ sin α = x
∴ – 2 sin^-1 x = – 2 α = cos^-1 (1 – x)
or cos 2α = 1 – x [∵ cos(- θ) = cos θ]
∴ 1 – 2 sin² α = (1 – x)
Putting sin α = x
⇒ 1 – 2x² = 1 – x
or 2x² – x = 0
x(2x – 1) = 0
∴ x = 0, $\frac{1}{2}$
But x = $\frac{1}{2}$ does not satisfy the equation.
∴ x = 0
Hence, the correct answer is (C).

Question 17.
tan^-1 $\left(\frac{x}{y}\right)$ – tan^-1 $\frac{x-y}{x+y}$ is equal to
(A) $\frac{\pi}{2}$

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

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

(D) $\frac{3 \pi}{4}$
Solution.
We have tan^-1 $\left(\frac{x}{y}\right)$ – tan^-1 $\frac{x-y}{x+y}$