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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

BITSAT 2024 · Paper 2 · Mathematics · Inverse Trigonometric Functions

If ( ^ -1 x)= \ ^ -1 ( a b^2-a^2 )\ b > a , then x=

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

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⁡\cot \left(\cos ^{-1} x\right)=\sec\left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}Since,Since,\cos ^{-1} x=\cot ^{-1} \left(\frac{x}{\sqrt{1-x^2}}\right)andtan⁡−1x=sec⁡−1(1+x2)⇒cot⁡(cot⁡−1x1−x2)=sec⁡{sec⁡−11+(ab2−a2)2}⇒x1−x2=b2−a2+a2b2−a2⇒x1−x2=bb2−a2and \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 \frac{x}{\sqrt{1-x^2}}=\sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}}\Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}}On squaring both the sides, we get ⇒ x21−x2=b2b2−a2\frac{x^2}{1-x^2}=\frac{b^2}{b^2-a^2}⇒x2b2−x2a2=b2−b2x2x2b2+b2x2−x2a2=b2\begin{aligned} \Rightarrow& x^2 b^2-x^2 a^2=b^2-b^2 x^2 \\ & x^2 b^2+b^2 x^2-x^2 a^2=b^2\end{aligned}⇒2x2b2−x2a2=b2⇒x2(2b2−a2)=b2⇒x=b2b2−a2\Rightarrow \quad 2 x^2 b^2-x^2 a^2=b^2 \Rightarrow x^2\left(2 b^2-a^2\right)=b^2\Rightarrow \quad x=\frac{b}{\sqrt{2 b^2-a^2}}
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