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

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Question 1.
cos^-1(cos$\frac { 13π }{ 6 }$)
Solution:
cos^-1(cos$\frac { 13π }{ 6 }$) = cos^-1cos(2π + $\frac { π }{ 6 }$ = cos^-1cos$\frac { π }{ 6 }$) = $\frac { π }{ 6 }$

Question 2.
tan^-1(tan$\frac { 7π }{ 6 }$)
Solution:
tan^-1(tan$\frac { 7π }{ 6 }$) = tan^-1tan(π + $\frac { π }{ 6 }$ = tan^-1tan$\frac { π }{ 6 }$) = $\frac { π }{ 6 }$.

Question 3.
sin^-1$\frac { 3 }{ 5 }$ = tan^-1$\frac { 24 }{ 7 }$
Solution:
Let sin^-1$\frac { 3 }{ 5 }$ = θ

Question 4.
sin^-1$\frac { 8 }{ 17 }$ + sin^-1$\frac { 3 }{ 5 }$ = sin^-1$\frac { 77 }{ 85 }$
Solution:

Question 5.
cos^-1$\frac { 4 }{ 5 }$ + cos^-1$\frac { 12 }{ 13 }$ = cos^-1$\frac { 33 }{ 65 }$
Solution:

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

Question 7.
tan^-1$\frac { 63 }{ 16 }$ = sin^-1$\frac { 5 }{ 13 }$ + cos^-1$\frac { 3 }{ 5 }$
Solution:

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

Question 9.
tan^-1$\sqrt{x}$ = $\frac { 1 }{ 2 }$cos^-1$\frac { 1-x }{ 1+x }$, x ∈ [0, 1]
Solution:
Put x = tan²θ. ∴ θ = tan^-1$\sqrt{x}$
R.H.S.
= $\frac { 1 }{ 2 }$cos^-1$\frac { 1-x }{ 1+x }$
= $\frac { 1 }{ 2 }$cos^-1$\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
= $\frac { 1 }{ 2 }$cos^-1(cos2θ) = $\frac { 1 }{ 2 }$ x 2θ = θ.
= tan^-1$\sqrt{x}$ = L.H.S
Hence, tan^-1$\sqrt{x}$ $\frac { 1 }{ 2 }$cos^-1$\frac { 1-x }{ 1+x }$

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

Question 11.
tan^-1$\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)$ = $\frac { π }{ 4 }$ + $\frac { 1 }{ 2 }$cos^-1, x ∈ (0, $\frac { π }{ 4 }$)
Solution:

Question 12.
$\frac { 9π }{ 8 }$ – $\frac { 9 }{ 4 }$sin^-1$\frac { 1 }{ 3 }$ = $\frac { 9 }{ 4 }$sin^-1$\frac{2 \sqrt{2}}{3}$
Solution:
$\frac { 9π }{ 8 }$ – $\frac { 9 }{ 4 }$sin^-1$\frac { 1 }{ 3 }$ = $\frac { 9 }{ 4 }$sin^-1$\frac{2 \sqrt{2}}{3}$
⇒ $\frac { 9 }{ 4 }$(sin^-1$\frac{2 \sqrt{2}}{3}$ + sin^-1$\frac { 1 }{ 3 }$) = $\frac { 9π }{ 8 }$

Hence $\frac { 9π }{ 8 }$ – $\frac { 9 }{ 4 }$sin^-1$\frac { 1 }{ 3 }$ = $\frac { 9 }{ 4 }$sin^-1$\frac{2 \sqrt{2}}{3}$

Question 13.
2tan^-1(cos x) = tan^-1(cosec x)
Solution:
Now L.H.S
= 2tan^-1(cos x)
= tan^-1$\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)$
= tan^-1$\left(\frac{2 \cos x}{\sin ^{2} x}\right)$
Putting this value in the given equation, we get
tan^-1$\left(\frac{2 \cos x}{\sin ^{2} x}\right)$ = tan^-1(2 cosec x)
⇒ $\frac{2 \cos x}{\sin ^{2} x}$ = 2 cosec x = $\frac { 2 }{ sin x }$
⇒ cos x = sin x or tan x = 1
⇒ x = $\frac { π }{ 4 }$

Question 14.
tan^-1$\frac { 1-x }{ 1+x }$ = $\frac { 1 }{ 2 }$tan^-1x, x > 0
Solution:

Question 15.
sin tan^-1x, |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^-1x = α ∴ tan α = x.
So, sin α = $\frac{x}{\sqrt{1+x^{2}}}$
or α = sin^-1$\frac{x}{\sqrt{1+x^{2}}}$
Now sin tan^-1x = sin α = sin(sin^-1$\frac{x}{\sqrt{1+x^{2}}}$)
= $\frac{x}{\sqrt{1+x^{2}}}$
∴ Part (D) is the correct answer.

Question 16.
If sin^-1(1-x)-2 sin^-1x = $\frac { π }{ 2 }$, then x is equal to
(A) 0, $\frac { 1 }{ 2 }$
(B) 1, $\frac { 1 }{ 2 }$
(C) 0
(D) $\frac { 1 }{ 2 }$
Solution:
sin^-1(1-x)-2 sin^-1x = $\frac { π }{ 2 }$
Putting $\frac { π }{ 2 }$ = sin^-1(1 – x) + cos^-1(1 – x),
sin^-1(1 – x) – 2 sin^-1 = sin^-1(1 – x) + cos^-1(1 – x)
⇒ – 2 sin^-1x = cos^-1(1 – x)
Let sin^-1x = α. ∴ sin a = α
∴ – 2 sin^-1x = – 2α = cos^-1(1 – x)
or cos 2α = 1 – x [∵ cos (-θ) = cos θ]
∴ 1 – 2sin²α = (1 – x)
Putting sin α = x, we get
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.
∴ Part (C) is the correct answer.

Question 17.
tan^-1($\frac { x }{ y }$) tan^-1($\frac { x-y }{ x+y }$) is equal to
(A) $\frac { π }{ 2 }$
(B) $\frac { π }{ 3 }$
(C) $\frac { π }{ 4 }$
(D) $\frac { -3π }{ 4 }$
Solution:
tan^-1($\frac { x }{ y }$) tan^-1($\frac { x-y }{ x+y }$)
Applying the formula tan^-1a tan^-1b = tan^-1$\frac { a-b }{ 1+ab }$, we get:
Given expression

∴ Part (C) is the required answer.