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tan-inverse MCQ Questions for Class 12

tan-inverse se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

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8 questions tagged with tan-inverse.

Question 1/8 Medium Mathematics 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 \(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

\(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).

Step 2

Why this answer is correct

The correct answer is A. यह सभी \(x\in\mathbb{R}\) के लिए सत्य है / It is true for all \(x\in\mathbb{R}\). \(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).

Step 3

Exam Tip

\(\frac{x}{\sqrt{1+x^2}}\) हमेशा ([-1,1]) में रहता है। इसलिए identity सभी real (x) के लिए मान्य है।

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Question 2/8 Medium Mathematics 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. \(\frac{5}{13}\)

Step 1

Concept

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

Step 2

Why this answer is correct

The correct answer is A. \(\frac{5}{13}\). From \(\tan\theta=\frac{5}{12}\), opposite is (5), adjacent is (12), and hypotenuse is (13). Hence \(\sin\theta=\frac{5}{13}\).

Step 3

Exam Tip

\(\tan\theta=\frac{5}{12}\) से opposite (5), adjacent (12), hypotenuse (13) है। इसलिए \(\sin\theta=\frac{5}{13}\)।

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Question 3/8 Medium Mathematics Inverse Trigonometric Functions Introduction to inverse trigonometric functions Class 12 Level 32

\(\tan^{-1}0\) का मुख्य मान क्या है?

What is the principal value of \(\tan^{-1}0\)?

Explanation opens after your attempt
Correct Answer

A. (0)

Step 1

Concept

\(\tan0=0\), so \(\tan^{-1}0=0\). Even in zero-value questions, check the principal range.

Step 2

Why this answer is correct

The correct answer is A. (0). \(\tan0=0\), so \(\tan^{-1}0=0\). Even in zero-value questions, check the principal range.

Step 3

Exam Tip

\(\tan0=0\), इसलिए \(\tan^{-1}0=0\)। zero value वाले प्रश्नों में भी principal range जांचें।

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Question 4/8 Medium Mathematics Inverse Trigonometric Functions Class 12 Level 32

\(\tan^{-1}(x^2+1)\) का डोमेन क्या है?

What is the domain of \(\tan^{-1}(x^2+1)\)?

Explanation opens after your attempt
Correct Answer

A. \(\mathbb{R}\)

Step 1

Concept

\(\tan^{-1}u\) is defined for every real (u). Hence \(x^2+1\) puts no restriction on (x).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}\). \(\tan^{-1}u\) is defined for every real (u). Hence \(x^2+1\) puts no restriction on (x).

Step 3

Exam Tip

\(\tan^{-1}u\) हर real \(u\) के लिए परिभाषित होता है। इसलिए \(x^2+1\) के कारण (x) पर कोई प्रतिबंध नहीं है।

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Question 5/8 Medium Mathematics Inverse Trigonometric Functions Class 12 Level 32

यदि \(\tan^{-1}x=-\frac{\pi}{3}\), तो \(x\) का मान क्या है?

If \(\tan^{-1}x=-\frac{\pi}{3}\), what is the value of \(x\)?

Explanation opens after your attempt
Correct Answer

A. -\(\sqrt{3}\)

Step 1

Concept

\(\tan^{-1}x=-\frac{\pi}{3}\) means \(x=\tan\left(-\frac{\pi}{3}\right)\). Hence \(x=-\sqrt{3}\).

Step 2

Why this answer is correct

The correct answer is A. -\(\sqrt{3}\). \(\tan^{-1}x=-\frac{\pi}{3}\) means \(x=\tan\left(-\frac{\pi}{3}\right)\). Hence \(x=-\sqrt{3}\).

Step 3

Exam Tip

\(\tan^{-1}x=-\frac{\pi}{3}\) का अर्थ है \(x=\tan\left(-\frac{\pi}{3}\right)\)। इसलिए \(x=-\sqrt{3}\)।

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Question 6/8 Medium Mathematics 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. \(\frac{x}{\sqrt{1+x^2}}\)

Step 1

Concept

If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{x}{\sqrt{1+x^2}}\). If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).

Step 3

Exam Tip

यदि \(\theta=\tan^{-1}x\), तो \(\tan\theta=x\) और hypotenuse \(\sqrt{1+x^2}\) है। इसलिए \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\)।

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Question 7/8 Medium Mathematics Inverse Trigonometric Functions Class 12 Level 32

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

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

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

\(\tan^{-1}x\) is also an odd function. Therefore a negative input changes the sign of the output.

Step 2

Why this answer is correct

The correct answer is A. \(-\tan^{-1}x\). \(\tan^{-1}x\) is also an odd function. Therefore a negative input changes the sign of the output.

Step 3

Exam Tip

\(\tan^{-1}x\) भी विषम function है। इसलिए negative input पर output का sign बदलता है।

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Question 8/8 Medium Mathematics Inverse Trigonometric Functions Class 12 Level 32

\(\tan^{-1}(\tan\theta)=\theta\) कब निश्चित रूप से सत्य है?

When is \(\tan^{-1}(\tan\theta)=\theta\) definitely true?

Explanation opens after your attempt
Correct Answer

B. \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\)

Step 1

Concept

The principal range of \(\tan^{-1}x\) is (\(-\frac{\pi}{2},\frac{\pi}{2}\)). Since \(\tan\) is periodic, the principal interval is necessary.

Step 2

Why this answer is correct

The correct answer is B. \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\). The principal range of \(\tan^{-1}x\) is (\(-\frac{\pi}{2},\frac{\pi}{2}\)). Since \(\tan\) is periodic, the principal interval is necessary.

Step 3

Exam Tip

\(\tan^{-1}x\) का principal range \((-\frac{\pi}{2},\frac{\pi}{2})\) है। \(\tan\) period वाला है, इसलिए principal interval जरूरी है।

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