Mathematics Questions and Answers – Properties of Inverse Trigonometric Functions

«
»

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.
advertisement

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⁡(\(\frac{5}{\sqrt{41}}\)))
=\(\frac{5}{\sqrt{41}}\)
advertisement
advertisement

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⁡{\(x\sqrt{1+y^2}-y \sqrt{1-x^2}\)}.
advertisement

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⁡\((\frac{7}{9})\).
advertisement

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⁡(\(\sqrt{1+x^2}\))
advertisement

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.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter