Q54 / 100 · BITSAT 2024 (Memory Based Paper 2) · Mathematics
If $\cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}$
$b \gt a$, then $\mathrm{x}=$
- Correct answerA$\frac{b}{\sqrt{2 b^2-a^2}}$
- B$\frac{\sqrt{b^2-a^2}}{a b}$
- C$\frac{a}{\sqrt{2 b^2-a^2}}$
- D$\frac{\sqrt{b^2-a^2}}{a}$
Solution
Given that,
$\cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}$
Since, $\cos ^{-1} x=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and $\tan ^{-1} x=\sec ^{-1}\left(\sqrt{1+x^2}\right)$
$\Rightarrow \cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right)=$
$\sec \left\{\sec ^{-1} \sqrt{1+\left(\frac{a}{\sqrt{b^2-a^2}}\right)^2}\right\}$
$\Rightarrow \frac{x}{\sqrt{1-x^2}}=\sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}}$
$\Rightarrow \frac{x}{\sqrt{1-x^2}}=\frac{b}{\sqrt{b^2-a^2}}$
On squaring both the sides, we get
$\Rightarrow \frac{x^2}{1-x^2}=\frac{b^2}{b^2-a^2}$
$\begin{aligned} \Rightarrow & x^2 b^2-x^2 a^2=b^2-b^2 x^2 \\ & x^2 b^2+b^2 x^2-x^2 a^2=b^2\end{aligned}$
$\Rightarrow \quad 2 x^2 b^2-x^2 a^2=b^2 \Rightarrow x^2\left(2 b^2-a^2\right)=b^2$
$\Rightarrow \quad x=\frac{b}{\sqrt{2 b^2-a^2}}$
Hence, the correct answer is option: 'A'.