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Mathematics Properties of inverse trigonometric functions MCQ Questions for Class 12 General

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01 Practice Topic Domain and range 24 MCQs 02 Practice Topic Graphs of inverse trigonometric functions 3 MCQs 03 Practice Topic Introduction to inverse trigonometric functions 106 MCQs 04 Practice Topic Principal value branch 69 MCQs

Properties of inverse trigonometric functions Practice Questions

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Question 31/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 31

\(\sin^{-1}\left(\frac{2}{3}\right)+\cos^{-1}\left(\frac{2}{3}\right)\) का मान क्या है?

What is the value of \(\sin^{-1}\left(\frac{2}{3}\right)+\cos^{-1}\left(\frac{2}{3}\right)\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{\pi}{2}\)

Explanation

Simple Explanation

हर \(x\in\left[-1,1\right]\) के लिए \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\)। यहां \(x=\frac{2}{3}\) मान्य है। / For every \(x\in\left[-1,1\right]\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{2}{3}\) is valid.

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Question 32/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 31

यदि \(\theta=\cot^{-1}\left(\frac{3}{4}\right)\), तो \(\cos\theta\) का मान क्या है?

If \(\theta=\cot^{-1}\left(\frac{3}{4}\right)\), what is the value of \(\cos\theta\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{3}{5}\)

Explanation

Simple Explanation

\(\cot\theta=\frac{3}{4}\) से आधार (3), लम्ब (4) और कर्ण (5) होगा। इसलिए \(\cos\theta=\frac{3}{5}\)। / From \(\cot\theta=\frac{3}{4}\), the base is (3), perpendicular is (4), and hypotenuse is (5). Hence \(\cos\theta=\frac{3}{5}\).

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Question 33/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 31

\(\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)\) का मान क्या है?

What is the value of \(\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{1}{3}\right)\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{\pi}{4}\)

Explanation

Simple Explanation

सूत्र से योग \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\) है। \(\tan^{-1}\) जोड़ते समय \(ab<1\) जांचें। / By the formula, the sum is \(\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan^{-1}1=\frac{\pi}{4}\). While adding \(\tan^{-1}\) terms, check \(ab<1\).

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Question 34/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 31

\(\tan\left(\cos^{-1}\frac{5}{13}\right)\) का मान क्या है?

What is the value of \(\tan\left(\cos^{-1}\frac{5}{13}\right)\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{12}{5}\)

Explanation

Simple Explanation

मान लें \(\theta=\cos^{-1}\frac{5}{13}\), तब \(\cos\theta=\frac{5}{13}\) और \(\sin\theta=\frac{12}{13}\)। इसलिए \(\tan\theta=\frac{12}{5}\)। / Let \(\theta=\cos^{-1}\frac{5}{13}\), then \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\). Therefore \(\tan\theta=\frac{12}{5}\).

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Question 35/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\tan^{-1}(-x)\) का सही रूप कौन-सा है?

Which is the correct form of \(\tan^{-1}(-x)\)?

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Correct Answer

A. \(-\tan^{-1}x\)

Explanation

Simple Explanation

\(\tan^{-1}x\) भी विषम function है। इसलिए negative input पर output का sign बदलता है। / \(\tan^{-1}x\) is also an odd function. Therefore a negative input changes the sign of the output.

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Question 36/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\cos^{-1}(-x)\) के लिए सही identity कौन-सी है?

Which identity is correct for \(\cos^{-1}(-x)\)?

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Correct Answer

B. \(\pi-\cos^{-1}x\)

Explanation

Simple Explanation

\(\cos^{-1}(-x)=\pi-\cos^{-1}x\) क्योंकि output \([0,\pi]\) में रहता है। इसे odd function न समझें। / \(\cos^{-1}(-x)=\pi-\cos^{-1}x\) because the output remains in \([0,\pi]\). Do not treat it as an odd function.

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Question 37/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\sin(\cos^{-1}x)\) का सही मान क्या है, जब \(x\in[-1,1]\)?

What is the correct value of \(\sin(\cos^{-1}x)\) when \(x\in[-1,1]\)?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{1-x^2}\)

Explanation

Simple Explanation

यदि \(\theta=\cos^{-1}x\), तो \(\theta\in[0,\pi]\) और \(\sin\theta\ge0\)। इसलिए मान \(\sqrt{1-x^2}\) है। / If \(\theta=\cos^{-1}x\), then \(\theta\in[0,\pi]\) and \(\sin\theta\ge0\). Hence the value is \(\sqrt{1-x^2}\).

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Question 38/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\cos(\sin^{-1}x)\) का सही मान क्या है, जब \(x\in[-1,1]\)?

What is the correct value of \(\cos(\sin^{-1}x)\) when \(x\in[-1,1]\)?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{1-x^2}\)

Explanation

Simple Explanation

यदि \(\theta=\sin^{-1}x\), तो \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) और \(\cos\theta\ge0\)। इसलिए positive root लेना चाहिए। / If \(\theta=\sin^{-1}x\), then \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\cos\theta\ge0\). So the positive root should be taken.

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Question 39/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\tan(\sin^{-1}x)\) का सही मान क्या है?

What is the correct value of \(\tan(\sin^{-1}x)\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{x}{\sqrt{1-x^2}}\)

Explanation

Simple Explanation

मान लें \(\theta=\sin^{-1}x\), तब \(\sin\theta=x\) और \(\cos\theta=\sqrt{1-x^2}\)। इसलिए \(\tan\theta=\frac{x}{\sqrt{1-x^2}}\)। / Let \(\theta=\sin^{-1}x\), then \(\sin\theta=x\) and \(\cos\theta=\sqrt{1-x^2}\). Hence \(\tan\theta=\frac{x}{\sqrt{1-x^2}}\).

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Question 40/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\sin(\tan^{-1}x)\) का सही मान क्या है?

What is the correct value of \(\sin(\tan^{-1}x)\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{x}{\sqrt{1+x^2}}\)

Explanation

Simple Explanation

यदि \(\theta=\tan^{-1}x\), तो \(\tan\theta=x\) और hypotenuse \(\sqrt{1+x^2}\) है। इसलिए \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\)। / If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).

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