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

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 ∈ $(0,\frac{\pi }{4})$
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.

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