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

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 

Question 1.
Find the value of cos^-1\(\left(\cos \frac{13 \pi}{6}\right)\).
Answer:
Principal value branch of cos^-1 is (0, π).

Question 2.
Find the value of tan^-1\(\left(\tan \frac{7 \pi}{6}\right)\).
Answer:

Prove that:

Question 3.
sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{24}{7}\right)\)
Answer:

Question 4.
Prove that:
sin^-1\(\left(\frac{8}{17}\right)\) + sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{77}{36}\right)\)
Answer:

Question 5.
Prove that:
cos^-1\(\frac{4}{5}\) + cos^-1\(\frac{12}{13}\) = cos^-1\(\frac{33}{65}\)
Answer:

Question 6.
cos^-1\(\frac{12}{13}\) + sin^-1\(\frac{3}{5}\) = sin^-1\(\frac{56}{65}\)
Answer:

Question 7.
Prove that
tan^-1\(\frac{63}{16}\) = sin^-1\(\frac{5}{13}\) + cos^-1\(\frac{3}{5}\)
Answer:

Question 8.
Prove that:
tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Answer:

Question 9.
Prove that:
tan^-1√x = \(\frac{1}{2}\) cos^-1\(\left(\frac{1-x}{1+x}\right)\), x ∈ [0, 1]
Answer:

Question 10.
Prove that
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 ∈ \(\left(0, \frac{\pi}{4}\right)\)
Answer:

Question 11.
Prove that:
tan^-1 \(\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]\) = \(\frac{\pi}{4}-\frac{1}{2}\) cos^-1x, - \(\frac{1}{\sqrt{2}}\) ≤ x ≤ 1
Answer:
Let x = cos 2θ

Thus, L.H.S. = R.H.S.
Hence Proved

Question 12.
Prove that:
\(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1 \(\left(\frac{1}{3}\right)\) = \(\frac{9}{4}\) sin^-1\(\left(\frac{2 \sqrt{2}}{3}\right)\)
Answer:

Solve the following equations:

Question 13.
tan^-1 (cos x) = tan^-1 (2 cosec x)
Answer:

⇒ cos x = sin²x × cosec x
⇒ cos x = sin x
⇒ tan x = 1 ⇒ tan x = tan \(\frac{\pi}{4}\)
Thus, x = \(\frac{\pi}{4}\)

Question 14.
tan^-1 \(\frac{1-x}{1+x}\) = \(\frac{1}{2}\) tan^-1x, (x > 0)
Answer:

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) \(\frac{x}{\sqrt{1+x^{2}}}\)
Answer:
Let tan^-1 x = θ ⇒ tan θ = x
∴ sin θ = tan θ = \(\frac{x}{1}\)

From ∆ABC, sin θ = \(\frac{x}{\sqrt{1+x^{2}}}\)
⇒ sin (tan^-1 x) = \(\frac{x}{\sqrt{1+x^{2}}}\)
Thus, option (D) is correct.

Question 16.
If sin^-1 (1 - x) - 2 sin^-1x = \(\frac{\pi}{2}\), then x is equal to:
(A) 0, \(\frac{1}{2}\)
(B) 1, \(\frac{1}{2}\)
(C) 0
(D) \(\frac{1}{2}\)
Answer:
sin^-1 (1 - x) - 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ sin^-1 (1 - x) - 2 sin^-1 x = sin^-1 (1 - x) + cos^-1 (1 - x)
⇒ - 2 sin^-1 x = cos^-1 (1 - x)
[∵ sin^-1 (1 - x) + cos^-1 (1 - x) = \(\frac{\pi}{2}\)]
Let sin^-1 x = θ
then sin θ = x
Let sin^-1 x = θ
⇒ sin θ = x
∴ - 2θ = cos^-1 (1 - x)
⇒ cos (- 2 θ) = 1 - x
⇒ cos 2θ = 1 - x [∵ cos (- θ) = cos θ] ...(i)
Since, we know that
cos 2 θ = 1 - 2 sin² θ
⇒ cos 2θ = 1 - 2x² ..... (ii)
From equations (i) and (ii), we have
1 - x = 1 - 2x²
⇒ 2x² - ² = 0
⇒ x (2x - 1) = 0
⇒ x = 0, x = \(\frac{1}{2}\)
But x = \(\frac{1}{2}\) does not satisfies given equation.
∴ x = 0
Thus, option (C) is correct.

Question 17.
tan^-1\(\left(\frac{x}{y}\right)\) - tan^-1\(\left(\frac{x-y}{x+y}\right)\) is equal to:
(A) \(\frac{\pi}{2}\)
(B) \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{4}\)
(D) \(\frac{3 \pi}{4}\)
Answer:

Thus, option (C) is correct.