Mathematics Questions and Answers – Properties of Inverse Trigonometric Functions

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This set of Mathematics 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\(\frac{⁡1}{x}\)
View Answer

Answer: b
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.
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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⁡\(\sqrt{x^2-1}\)
View Answer

Answer: d
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) \(\frac{4}{5}\)
View Answer

Answer: b
Explanation: From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q3
tan-1⁡(\(\frac{4}{5}\))=cos-1⁡(\(\frac{5}{\sqrt{41}}\))
cos⁡(tan-1⁡(\(\frac{4}{5}\))=cos⁡(cos-1⁡(\(\frac{5}{\sqrt{41}}\)))
=\(\frac{5}{\sqrt{41}}\)
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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) \(\sqrt{1+x^2}\)
View Answer

Answer: a
Explanation: Let sin-1⁡x=y. From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q4
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)
View Answer

Answer: b
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⁡(\(\frac{64}{65}\))
View Answer

Answer: a
Explanation: From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q6
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
View Answer

Answer: c
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⁡\((\frac{7}{9})\).
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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) \(\frac{π}{4}\)
View Answer

Answer: b
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⁡\((\sqrt{1+x^2})\)
View Answer

Answer: c
Explanation: Let 2 tan-1⁡x=y
⇒tan-1⁡x=\(\frac{y}{2}\)
From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q9
⇒tan-1⁡x=sec-1⁡\(\sqrt{1+x^2}=\frac{y}{2}\) ⇒y=2 sec-1⁡(\(\sqrt{1+x^2}\))
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10. sin-1⁡x+cos1⁡x= ___
a) \(\frac{π}{2}\)
b) π
c) \(\frac{π}{3}\)
d) 2π
View Answer

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

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

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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