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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 41/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

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

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

Explanation opens after your attempt
Correct Answer

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

Explanation

Simple Explanation

\(\theta=\tan^{-1}x\) लेने पर adjacent side (1) और hypotenuse \(\sqrt{1+x^2}\) मिलती है। इसलिए \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\)। / Taking \(\theta=\tan^{-1}x\), the adjacent side is (1) and the hypotenuse is \(\sqrt{1+x^2}\). Thus \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\).

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Question 42/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

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

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

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Correct Answer

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

Explanation

Simple Explanation

यह सीधे identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) से मिलता है। यहाँ \(x=\frac{\sqrt{3}}{2}\) valid domain में है। / This follows directly from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{\sqrt{3}}{2}\) is in the valid domain.

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Question 43/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

यदि \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), तो \(\cos\theta\) का मान क्या है?

If \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), what is the value of \(\cos\theta\)?

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Correct Answer

A. \(\frac{4}{5}\)

Explanation

Simple Explanation

\(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\) है, इसलिए \(\cos\theta\) positive होगा। (3,4,5) triangle से \(\cos\theta=\frac{4}{5}\)। / \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\cos\theta\) is positive. From the (3,4,5) triangle, \(\cos\theta=\frac{4}{5}\).

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Question 44/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

यदि \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\), तो \(\sin\theta\) क्या होगा?

If \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\), what is \(\sin\theta\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{5}{13}\)

Explanation

Simple Explanation

\(\tan\theta=\frac{5}{12}\) से opposite (5), adjacent (12), hypotenuse (13) है। इसलिए \(\sin\theta=\frac{5}{13}\)। / From \(\tan\theta=\frac{5}{12}\), opposite is (5), adjacent is (12), and hypotenuse is (13). Hence \(\sin\theta=\frac{5}{13}\).

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Question 45/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

f(x)=\cos^{-1}x अपने domain \([-1,1]\) पर किस प्रकार का फलन है?

What type of function is f(x)=\cos^{-1}x on its domain \([-1,1]\)?

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Correct Answer

B. घटतीdecreasing

Explanation

Simple Explanation

f(x)=\cos^{-1}x के लिए \\( \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}<0 \\) for x\in(-1,1). अवकलज नकारात्मक होने से फलन घटता है। सीमाओं पर भी \(\cos^{-1}(-1)=\pi\) और \(\cos^{-1}(1)=0\) आता है, अतः पूरे बंद अंतराल \([-1,1]\) पर यह घटता क्रम में है। विकल्प 'बढ़ती' गलत है क्योंकि अवकलज का चिन्ह नकारात्मक है; 'स्थिर' व 'आवर्त' भी उपयुक्त नहीं हैं। परीक्षा सुझाव: तुरंत monotonicity जाँचने के लिए अवकलज ज्ञात करें या याद रखें कि \(\cos\theta\) मुख्य-श्रेणी \([0,\pi]\) पर घटता है, अतः inverse घटेगा। / For f(x)=\cos^{-1}x, \\( \frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}<0 \\) for x in (-1,1). A negative derivative implies the function is decreasing. At endpoints, \(\cos^{-1}(-1)=\pi\) and \(\cos^{-1}(1)=0\), so the function decreases over the whole closed interval [-1,1]. 'Increasing' is wrong because the derivative is negative; 'constant' and 'periodic' are also not applicable. Exam tip: test monotonicity by checking the sign of the derivative or recall that \(\cos\theta\) is decreasing on [0,\pi], so its inverse is decreasing on [-1,1].

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Question 46/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\) के लिए सही कथन कौन-सा है?

Which statement is correct for \(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)?

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Correct Answer

A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य हैIt is true for all \(x\in\mathbb{R}\)

Explanation

Simple Explanation

\(\frac{x}{\sqrt{1+x^2}}\) हमेशा ([-1,1]) में रहता है। इसलिए identity सभी real (x) के लिए मान्य है। / \(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).

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Question 47/48 Medium Mathematics Inverse Trigonometric Functions Properties of inverse trigonometric functions Class 12 Level 32

\(\sin\left(\cos^{-1}\left(\frac{7}{25}\right)\right)\) का मान क्या है?

What is the value of \(\sin\left(\cos^{-1}\left(\frac{7}{25}\right)\right)\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{24}{25}\)

Explanation

Simple Explanation

\(\theta=\cos^{-1}\left(\frac{7}{25}\right)\) लेने पर \(\theta\in[0,\pi]\) और \(sin\theta\ge0\)। इसलिए (7,24,25) triangle से \(\sin\theta=\frac{24}{25}\)। / Taking \(\theta=\cos^{-1}\left(\frac{7}{25}\right)\), we have \(\theta\in[0,\pi]\) and \(sin\theta\ge0\). Hence from the (7,24,25) triangle, \(\sin\theta=\frac{24}{25}\).

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

\(\tan\left(\cos^{-1}x\right)\) का सही मान क्या है, जब \(0<x<1\)?

What is the correct value of \(\tan\left(\cos^{-1}x\right)\), when \(0<x<1\)?

Explanation opens after your attempt
Correct Answer

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

Explanation

Simple Explanation

यदि \(\theta=\cos^{-1}x\), तो \(\cos\theta=x\) और \(\sin\theta=\sqrt{1-x^2}\)। इसलिए \(\tan\theta=\frac{\sqrt{1-x^2}}{x}\)। / If \(\theta=\cos^{-1}x\), then \(\cos\theta=x\) and \(\sin\theta=\sqrt{1-x^2}\). Hence \(\tan\theta=\frac{\sqrt{1-x^2}}{x}\).

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