# 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$$\frac { 13Ï€ }{ 6 }$$)
Solution:
cos-1(cos$$\frac { 13Ï€ }{ 6 }$$) = cos-1cos(2Ï€ + $$\frac { Ï€ }{ 6 }$$ = cos-1cos$$\frac { Ï€ }{ 6 }$$) = $$\frac { Ï€ }{ 6 }$$

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

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

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$$\frac { 1 }{ 8 }$$ = $$\frac { Ï€ }{ 4 }$$
Solution:

Question 9.
tan-1$$\sqrt{x}$$ = $$\frac { 1 }{ 2 }$$cos-1$$\frac { 1-x }{ 1+x }$$, x âˆˆ [0, 1]
Solution:
Put x = tanÂ²Î¸. âˆ´ Î¸ = 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(cos2Î¸) = $$\frac { 1 }{ 2 }$$ x 2Î¸ = Î¸.
= 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 âˆˆ (0, $$\frac { Ï€ }{ 4 }$$)
Solution:

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

Question 12.
$$\frac { 9Ï€ }{ 8 }$$ – $$\frac { 9 }{ 4 }$$sin-1$$\frac { 1 }{ 3 }$$ = $$\frac { 9 }{ 4 }$$sin-1$$\frac{2 \sqrt{2}}{3}$$
Solution:
$$\frac { 9Ï€ }{ 8 }$$ – $$\frac { 9 }{ 4 }$$sin-1$$\frac { 1 }{ 3 }$$ = $$\frac { 9 }{ 4 }$$sin-1$$\frac{2 \sqrt{2}}{3}$$
â‡’ $$\frac { 9 }{ 4 }$$(sin-1$$\frac{2 \sqrt{2}}{3}$$ + sin-1$$\frac { 1 }{ 3 }$$) = $$\frac { 9Ï€ }{ 8 }$$

Hence $$\frac { 9Ï€ }{ 8 }$$ – $$\frac { 9 }{ 4 }$$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)
â‡’ $$\frac{2 \cos x}{\sin ^{2} x}$$ = 2 cosec x = $$\frac { 2 }{ sin x }$$
â‡’ cos x = sin x or tan x = 1
â‡’ x = $$\frac { Ï€ }{ 4 }$$

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 = Î± âˆ´ tan Î± = x.
So, sin Î± = $$\frac{x}{\sqrt{1+x^{2}}}$$
or Î± = sin-1$$\frac{x}{\sqrt{1+x^{2}}}$$
Now sin tan-1x = sin Î± = sin(sin-1$$\frac{x}{\sqrt{1+x^{2}}}$$)
= $$\frac{x}{\sqrt{1+x^{2}}}$$
âˆ´ Part (D) is the correct answer.

Question 16.
If sin-1(1-x)-2 sin-1x = $$\frac { Ï€ }{ 2 }$$, 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 = $$\frac { Ï€ }{ 2 }$$
Putting $$\frac { Ï€ }{ 2 }$$ = sin-1(1 – x) + cos-1(1 – x),
sin-1(1 – x) – 2 sin-1 = sin-1(1 – x) + cos-1(1 – x)
â‡’ – 2 sin-1x = cos-1(1 – x)
Let sin-1x = Î±. âˆ´ sin a = Î±
âˆ´ – 2 sin-1x = – 2Î± = cos-1(1 – x)
or cos 2Î± = 1 – x [âˆµ cos (-Î¸) = cos Î¸]
âˆ´ 1 – 2sinÂ²Î± = (1 – x)
Putting sin Î± = x, we get
1 – 2xÂ² = 1 – x
or 2xÂ² – x = 0
â‡’ x(2x – 1) = 0. âˆ´ x = 0, $$\frac { 1 }{ 2 }$$
But x = $$\frac { 1 }{ 2 }$$ does not satisfy the equation. âˆ´ x = 0.
âˆ´ Part (C) is the correct answer.

Question 17.
tan-1($$\frac { x }{ y }$$) tan-1($$\frac { x-y }{ x+y }$$) is equal to
(A) $$\frac { Ï€ }{ 2 }$$
(B) $$\frac { Ï€ }{ 3 }$$
(C) $$\frac { Ï€ }{ 4 }$$
(D) $$\frac { -3Ï€ }{ 4 }$$
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

âˆ´ Part (C) is the required answer.