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

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
cos-1(cos$$\frac { 13π }{ 6 }$$)
Solution:
The principal value brach of cos-1 is [0, p]

Question 2.
tan-1(tan$$\frac { 7π }{ 6 }$$)
Solution:
The principal value brach of tan-1 is (- $$\frac { π }{ 2 }$$, $$\frac { π }{ 2 }$$)

Question 3.
2sin-1$$\frac { 3 }{ 5 }$$ = tan-1$$\frac { 24 }{ 7 }$$
Solution:

Question 4.
sin-1$$\frac { 8 }{ 17 }$$ + sin-1$$\frac { 3 }{ 5 }$$ = tan-1$$\frac { 77 }{ 36 }$$
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$$\left(\frac{1-x}{1+x}\right), x \in[0,1]$$
Solution:

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, $$\frac { π }{ 4 }$$)
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, – $$\frac{1}{\sqrt{2}} \leq x \leq 1$$
Solution:

Question 12.
$$\frac { 9p }{ 8 }$$ – $$\frac { 9 }{ 4 }$$sin-1$$\frac { 1 }{ 3 }$$ = $$\frac { 9 }{ 4 }$$sin-1$$\frac{2 \sqrt{2}}{3}$$
Solution:

Question 13.
2tan-1(cosx) = tan-1(2cosecx)
Solution:

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:
d. $$\frac{x}{\sqrt{1+x^{2}}}$$

Question 16.
sin-1(1 – x) – 2sin-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:
c. 0

But x ≠ $$\frac { 1 }{ 2 }$$, since sin-1(1 – $$\frac { 1 }{ 2 }$$) ≠ $$\frac { π }{ 2 }$$ + 2 sin-1$$\frac { 1 }{ 2 }$$
∴ x = 0.

Question 17.
tan-1 $$\frac { x }{ y }$$ – tan-1 $$\frac{x-y}{x+y}$$
Solution:
a. $$\frac { π }{ 2 }$$
b. $$\frac { π }{ 3 }$$
c. $$\frac { π }{ 4 }$$
d. $$\frac { -3π }{ 4 }$$
Solution:
c. $$\frac { π }{ 4 }$$

error: Content is protected !!