What is the correct value of \(\tan(\sin^{-1}x)\)?
Correct answer: A
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}}).
View question detailsMathematics
Properties of inverse trigonometric functions
प्रतिलोम त्रिकोणमितीय फलनों के गुणधर्म
In Class 12 Mathematics, this topic from the chapter “Inverse Trigonometric Functions” helps students understand the principal values, domains, and ranges of sin⁻¹x, cos⁻¹x, tan⁻¹x and related functions. It covers standard identities, relationships between inverse trigonometric functions, and the conditions needed to simplify expressions correctly. Students also learn to evaluate compositions and use these properties to solve equations and verify results without confusing restricted angles or function branches.
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Medium · Level 32 · inverse-trigonometric-functions,composition,tan-value
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- (\frac{x}{\sqrt{1-x^2}})
- (\frac{\sqrt{1-x^2}}{x})
- (\sqrt{1-x^2})
- (\frac{x}{\sqrt{1+x^2}})
Medium · Level 32 · inverse-trigonometric-functions,composition,tan-inverse
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- (\frac{x}{\sqrt{1+x^2}})
- (\frac{1}{\sqrt{1+x^2}})
- (\sqrt{1+x^2})
- (\frac{\sqrt{1+x^2}}{x})
Medium · Level 32 · inverse-trigonometric-functions,composition,cos-value
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- (\frac{1}{\sqrt{1+x^2}})
- (\frac{x}{\sqrt{1+x^2}})
- (\sqrt{1+x^2})
- (\frac{1}{1+x^2})
Medium · Level 32 · inverse-trigonometric-functions,identity,numerical
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- \(\frac{\pi}{2}\)
- \(\frac{\pi}{3}\)
- \(\frac{5\pi}{6}\)
- \(\pi\)
Medium · Level 32 · inverse-trigonometric-functions,triangle-method,composition
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- \(\frac{4}{5}\)
- \(-\frac{4}{5}\)
- \(\frac{3}{5}\)
- \(\frac{5}{4}\)
Medium · Level 32 · inverse-trigonometric-functions,triangle-method,tan-inverse
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- \(\frac{5}{13}\)
- \(\frac{12}{13}\)
- \(\frac{13}{5}\)
- \(\frac{5}{12}\)
Medium · Level 32 · inverse-trigonometry,complementary-identity,domain,Properties of inverse trigonometric functions,Inverse Trigonometric Functions,Mathematics,Class 12 MCQ
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- When (x\in[-1,1])
- When (x\in\mathbb{R})
- When (x>1)
- When (x<-1)
Medium · Level 32 · inverse-trigonometric-functions,monotonicity,arccos,derivative,class-12
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- increasing
- decreasing
- constant
- periodic
Medium · Level 32 · inverse-trigonometric-functions,identity,tan-inverse
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- It is true for all \(x\in\mathbb{R}\)
- It is true only for \(x\in[-1,1]\)
- It is never true
- It is true only for \(x>1\)
Medium · Level 32 · inverse-trigonometric-functions,triangle-method,cos-inverse
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- \(\frac{24}{25}\)
- \(\frac{7}{25}\)
- -\(\frac{24}{25}\)
- \(\frac{25}{24}\)
Medium · Level 32 · inverse-trigonometric-functions,composition,tan-value
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- \(\frac{\sqrt{1-x^2}}{x}\)
- \(\frac{x}{\sqrt{1-x^2}}\)
- \(\sqrt{1+x^2}\)
- -\(\frac{\sqrt{1-x^2}}{x}\)
What is the correct value of \(\sin(\tan^{-1}x)\)?
Correct answer: A
If \(\theta=\tan^{-1}x\), then \(\tan\theta=x\) and the hypotenuse is \(\sqrt{1+x^2}\). So \(\sin\theta=\frac{x}{\sqrt{1+x^2}}\).
View question detailsWhat is the correct value of \(\cos(\tan^{-1}x)\)?
Correct answer: A
Taking \(\theta=\tan^{-1}x\), the adjacent side is (1) and the hypotenuse is \(\sqrt{1+x^2}\). Thus \(\cos\theta=\frac{1}{\sqrt{1+x^2}}\).
View question detailsWhat is the value of \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)?
Correct answer: A
This follows directly from \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\). Here \(x=\frac{\sqrt{3}}{2}\) is in the valid domain.
View question detailsIf \(\theta=\sin^{-1}\left(\frac{3}{5}\right)\), what is the value of \(\cos\theta\)?
Correct answer: A
\(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\cos\theta\) is positive. From the (3,4,5) triangle, \(\cos\theta=\frac{4}{5}\).
View question detailsIf \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\), what is \(\sin\theta\)?
Correct answer: A
From \(\tan\theta=\frac{5}{12}\), opposite is (5), adjacent is (12), and hypotenuse is (13). Hence \(\sin\theta=\frac{5}{13}\).
View question detailsWhen is (\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x) true?
Correct answer: A
The governing complementary-angle identity is sin⁻¹x + cos⁻¹x = π/2. Rearranging gives cos⁻¹x = π/2 − sin⁻¹x. Both inverse functions must have real values, and the real domain of each is −1 ≤ x ≤ 1. Thus the identity is true precisely for x ∈ [−1, 1], making option A correct. The endpoints are included because sin⁻¹(±1) and cos⁻¹(±1) are defined. Option B is too broad: inverse sine and inverse cosine are not real for every real x. Options C and D lie outside the real domain and therefore cannot satisfy the stated real-valued identity.
View question detailsWhat type of function is f(x)=\cos^{-1}x on its domain \([-1,1]\)?
Correct answer: B
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].
Which statement is correct for \(\tan^{-1}x=\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\)?
Correct answer: A
\(\frac{x}{\sqrt{1+x^2}}\) always lies in ([-1,1]). Therefore the identity is valid for all real (x).
View question detailsWhat is the value of \(\sin\left(\cos^{-1}\left(\frac{7}{25}\right)\right)\)?
Correct answer: A
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, \(\sin\theta=\frac{24}{25}\).
View question detailsWhat is the correct value of \(\tan\left(\cos^{-1}x\right)\), when \(0<x<1\)?
Correct answer: A
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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