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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

aligned & If ( ^ -1 x)= \ ^ -1 ( a b^2-a^2 )\ : & b>a, then x=...

 If cot⁡(cos⁡−1x)=sec⁡{tan⁡−1(ab2−a2)}:b>a, then x=\begin{aligned} & \text { If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}: \\ & b>a, \text { then } \mathrm{x}=\end{aligned}

Options

  1. Ab2b2−a2\frac{b}{\sqrt{2 b^2-a^2}}
  2. Bb2−a2ab\frac{\sqrt{b^2-a^2}}{a b}
  3. Ca2b2−a2\frac{a}{\sqrt{2 b^2-a^2}}
  4. Db2−a2a\frac{\sqrt{b^2-a^2}}{a}

Correct answer

A. b2b2−a2\frac{b}{\sqrt{2 b^2-a^2}}

Step-by-step solution

Given that, cot⁡(cos⁡−1x)=sec⁡{tan⁡−1(ab2−a2)} Since, cos⁡−1x=cot⁡−1(x1−x2) and tan⁡−1x=sec⁡−1(1+x2)⇒cot⁡(cot⁡−1x1−x2)=sec⁡{sec⁡−11+(ab2−a2)2}⇒x1−x2=b2−a2+a2b2−a2⇒x1−x2=bb2−a2\begin{aligned} & \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & \text { Since, } \cos ^{-1} x=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \\ & \quad \text { and } \tan ^{-1} x=\sec ^{-1}\left(\sqrt{1+x^2}\right) \\ & \Rightarrow \cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right)=\sec \\ & \left\{\sec ^{-1} \sqrt{1+\left(\frac{a}{\sqrt{b^2-a^2}}\right)^2}\right\} \\ & \Rightarrow \quad \frac{x}{\sqrt{1-x^2}}=\sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}} \\ & \Rightarrow \quad \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}}\end{aligned} On squaring both the sides, we get x21−x2=b2b2−a2⇒x2b2−x2a2=b2−b2x2x2b2+b2x2−x2a2=b22x2b2−x2a2=b2⇒x2(2b2−a2)=b2⇒x=b2b2−a2\begin{aligned} & \frac{x^2}{1-x^2}=\frac{b^2}{b^2-a^2} \\ & \Rightarrow \quad x^2 b^2-x^2 a^2=b^2-b^2 x^2 \\ & \quad x^2 b^2+b^2 x^2-x^2 a^2=b^2 \\ & \quad 2 x^2 b^2-x^2 a^2=b^2 \\ & \Rightarrow \quad x^2\left(2 b^2-a^2\right)=b^2 \\ & \Rightarrow \quad x=\frac{b}{\sqrt{2 b^2-a^2}} \end{aligned}
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