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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If x=cosec ( ^ -1 ( ( ^ -1 ( ( ^ -1 a))))) , a ∈[ 0,1]

If x=\operatorname{cosec} \left(\tan ^{-1} \left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1}a\right)\right)\right)\right)\right),a∈[, \mathrm{a} \in[0,1]

Options

  1. Ax2−a2=3x^2-\mathrm{a}^2=3
  2. Bx2+a2=3x^2+\mathrm{a}^2=3
  3. Cx2−a2=2x^2-\mathrm{a}^2=2
  4. Dx2+a2=2x^2+\mathrm{a}^2=2

Correct answer

B. x2+a2=3x^2+\mathrm{a}^2=3

Step-by-step solution

x=cosec⁡(tan⁡−1(cos⁡(cot⁡−1(sec⁡(sin⁡−1a)))))=cosec⁡(tan⁡−1(cos⁡(cot⁡−1(sec⁡(sec⁡−111−a2)))))=cosec⁡(tan⁡−1(cos⁡(cot⁡−1(11−a2))))=cosec⁡(tan⁡−1(cos⁡(cos⁡−112−a2)))=cosec⁡(tan⁡−1(12−a2))=cosec⁡(cosec⁡−1(3−a2))∴x=3−a2∴x2+a2=3\begin{aligned} & x=\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sin ^{-1} \mathrm{a}\right)\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\sec \left(\sec ^{-1} \frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cot ^{-1}\left(\frac{1}{\sqrt{1-\mathrm{a}^2}}\right)\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\cos \left(\cos ^{-1} \frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right)\right) \\ & =\operatorname{cosec}\left(\tan ^{-1}\left(\frac{1}{\sqrt{2-\mathrm{a}^2}}\right)\right) \\ & =\operatorname{cosec}\left(\operatorname{cosec}^{-1}\left(\sqrt{3-\mathrm{a}^2}\right)\right) \\ \therefore \quad & x=\sqrt{3-\mathrm{a}^2} \\ \therefore \quad & x^2+\mathrm{a}^2=3\end{aligned}
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