# Class 12 Maths MCQ – Properties of Inverse Trigonometric Functions

This set of Class 12 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Inverse Trigonometric Functions”.

1. sin-1⁡x in terms of cos-1⁡is ____________
a) cos-1⁡$$\sqrt{1+x^2}$$
b) cos-1⁡$$\sqrt{1-x^2}$$
c) cos-1⁡x
d) cos-1$$\frac{⁡1}{x}$$

Explanation: Let sin-1⁡x=y
⇒x=sin⁡y
⇒$$x=\sqrt{1-cos^2⁡y}$$
⇒$$x^2=1-cos^2⁡y$$
⇒$$cos^2⁡y=1-x^2$$
∴y=cos-1⁡$$\sqrt{1-x^2}$$=sin-1⁡x.

2. What is sec-1⁡x in terms of tan-1⁡?
a) tan-1⁡$$\sqrt{1+x^2}$$
b) tan-1⁡1+x2
c) tan-1⁡x
d) tan-1⁡$$\sqrt{x^2-1}$$

Explanation: Let sec-1⁡x=y
⇒x=sec⁡y
⇒x=$$\sqrt{1+tan^2⁡y}$$
⇒x2-1=tan2⁡y
∴y=tan-1⁡$$\sqrt{x^2-1}$$=sec-1⁡x.

3. What is the value of cos⁡(tan-1⁡($$\frac{4}{5}$$))?
a) $$\frac{5}{4}$$
b) $$\frac{5}{\sqrt{41}}$$
c) $$\frac{\sqrt{41}}{5}$$
d) $$\frac{4}{5}$$

Explanation: From ∆ABC, we get

tan-1⁡($$\frac{4}{5}$$)=cos-1⁡($$\frac{5}{\sqrt{41}}$$)
cos⁡(tan-1⁡($$\frac{4}{5}$$)=cos⁡(cos-1⁡($$\frac{5}{\sqrt{41}}$$))
=$$\frac{5}{\sqrt{41}}$$

4. What is the solution of cot⁡(sin-1⁡x)?
a) $$\frac{\sqrt{1-x^2}}{x}$$
b) x
c) $$\sqrt{1-x^2}$$
d) $$\sqrt{1+x^2}$$

Explanation: Let sin-1⁡x=y. From ∆ABC, we get

y=sin-1⁡x=cot-1⁡($$\frac{\sqrt{1-x^2}}{x}$$)
∴cot⁡(sin-1⁡x)=$$cot⁡(cot^{-1}(\frac{\sqrt{1-x^2}}{x}))=\frac{\sqrt{1-x^2}}{x}$$.

5. Which of the following formula is incorrect?
a) sin-1⁡x+sin-1⁡y=sin-1⁡{$$x\sqrt{1-y^2}+y\sqrt{1-x^2}$$}
b) sin-1⁡x-sin-1⁡y=sin-1⁡{$$x\sqrt{1+y^2}+y\sqrt{1+x^2}$$}
c) 2 tan-1⁡x=tan-1⁡$$(\frac{2x}{1-x^2})$$
d) 2 cos-1⁡x=cos-1⁡(3x-4x3)

Explanation: The formula sin-1⁡x-sin-1⁡y=sin-1⁡{$$x\sqrt{1+y^2}+y\sqrt{1+x^2}$$} is incorrect. The correct formula is sin-1⁡x-sin-1⁡y=sin-1⁡{$$x\sqrt{1+y^2}-y \sqrt{1-x^2}$$}.
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6. Find the value of sin-1⁡($$\frac{5}{13}$$)+cos-1⁡($$\frac{3}{5}$$).
a) sin-1⁡($$\frac{63}{65}$$)
b) sin-1⁡1
c) 0
d) sin-1⁡($$\frac{64}{65}$$)

Explanation: From ∆ABC, we get

cos-1⁡$$(\frac{3}{5})$$=sin-1⁡$$(\frac{4}{5})$$
∴sin-1⁡($$\frac{5}{13}$$)+cos-1⁡($$\frac{3}{5}$$)=sin-1⁡($$\frac{5}{13}$$)+sin-1⁡($$\frac{4}{5}$$)
=sin-1⁡$$(\frac{5}{13}\sqrt{1-(\frac{4}{5})^2}+\frac{4}{5}\sqrt{1-(\frac{5}{13})^2})$$
=$$sin^{-1}(\frac{5}{13}×\frac{3}{5}+\frac{4}{5}×\frac{12}{13})=sin^{-1}(\frac{15+48}{65})=sin^{-1}(\frac{63}{65})$$.

7. Find the value of tan-1⁡($$\frac{1}{3}$$)+tan-1⁡($$\frac{1}{5}$$)+tan-1⁡(\frac{1}{7})[/latex]
a) tan-1⁡$$(\frac{4}{7})$$
b) tan-1⁡$$(\frac{9}{7})$$
c) tan-1⁡$$(\frac{7}{9})$$
d) tan-1⁡1

Explanation: Using the formula tan-1⁡x+tan-1⁡y=tan-1⁡$$\frac{x+y}{1-xy}$$, we get
tan-1⁡($$\frac{1}{3}$$)+tan-1⁡($$\frac{1}{5}$$)=tan-1⁡$$\bigg(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}×\frac{1}{5}}\bigg)$$
= $$tan^{-1}\bigg(\frac{\frac{8}{15}}{\frac{14}{15}}\bigg)=tan^{-1}⁡(\frac{8}{15}×\frac{15}{14})=tan^{-1}⁡(\frac{4}{7})$$
=$$tan^{-1}(\frac{1}{3})+tan^{-1}(\frac{1}{5})+tan^{-1}⁡(\frac{1}{7})=tan^{-1}(\frac{4}{7})+tan^{-1}⁡(\frac{1}{7})$$
=$$tan^{-1}⁡\bigg(\frac{\frac{4}{7} + \frac{1}{7}}{1-\frac{4}{7}×\frac{1}{7}}\bigg) = tan^{-1}⁡\bigg(\frac{\frac{5}{7}}{\frac{45}{49}}\bigg)=tan^{-1}⁡(\frac{5}{7}×\frac{49}{45})$$
=tan-1⁡$$(\frac{7}{9})$$.

8. Find the value of sin-1⁡($$\frac{3}{5}$$)+sin-1⁡($$\frac{4}{5}$$)+cos-1⁡$$(\frac{\sqrt{3}}{2})$$.
a) $$\frac{π}{3}$$
b) $$\frac{2π}{3}$$
c) $$\frac{4π}{3}$$
d) $$\frac{π}{4}$$

Explanation: Using the formula sin-1⁡x+sin-1⁡y=sin-1⁡$${x \sqrt{1-y^2}+y \sqrt{1-x^2}}$$, we get
sin-1⁡($$\frac{3}{5}$$)+sin-1⁡($$\frac{4}{5}$$)=sin-1$$\Big\{ \frac{3}{5} \sqrt{1-(\frac{4}{5})^2}+\frac{4}{5} \sqrt{1-(\frac{3}{5})^2} \Big\}$$
=sin-1⁡($$\frac{3}{5}$$×$$\frac{3}{5}$$+$$\frac{4}{5}$$×$$\frac{4}{5}$$)=sin-1⁡$$(\frac{25}{25})=\frac{π}{2}$$
∴ sin-1⁡($$\frac{3}{5}$$)+sin-1⁡($$\frac{4}{5}$$)+cos-1⁡$$(\frac{\sqrt{3}}{2})=\frac{π}{2}+\frac{π}{6}=\frac{3π+π}{6}=\frac{2π}{3}$$.

9. What is the value of 2 tan-1⁡x in terms of sin-1⁡?
a) sec-1⁡x
b) 2 sec-1⁡x
c) 2 sec-1⁡$$(\sqrt{1+x^2})$$
d) sec-1⁡$$(\sqrt{1+x^2})$$

Explanation: Let 2 tan-1⁡x=y
⇒tan-1⁡x=$$\frac{y}{2}$$
From ∆ABC, we get

⇒tan-1⁡x=sec-1⁡$$\sqrt{1+x^2}=\frac{y}{2}$$ ⇒y=2 sec-1⁡($$\sqrt{1+x^2}$$)

10. sin-1⁡x+cos1⁡x= ___
a) $$\frac{π}{2}$$
b) π
c) $$\frac{π}{3}$$
d) 2π

Explanation: sin-1⁡x+cos-1⁡x=$$\frac{π}{2}$$; x∈[-1,1]

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.