MCQ
If $\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}\right)$, then $x$ is equal to
- ✓$\frac{b}{\sqrt{2 b^2-a^2}}$
- B$\frac{a}{\sqrt{2 b^2-a^2}}$
- C$\frac{\sqrt{2 b^2-a^2}}{a}$
- D$\frac{\sqrt{2 b^2-a^2}}{b}$
✓
Answer
Correct option: A.
$\frac{b}{\sqrt{2 b^2-a^2}}$
(a) : We have,
$
\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}\right)
$
Let $\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}=\theta \Rightarrow \tan \theta=\frac{a}{\sqrt{b^2-a^2}}$
$\Rightarrow \sec \theta=\frac{b}{\sqrt{b^2-a^2}}$
$\therefore \quad \cot \left(\cos ^{-1} x\right)=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow \cos ^{-1} x=\cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)$
Again, let $\cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)=\phi \Rightarrow \cot \phi=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow \cos \phi=\frac{b}{\sqrt{2 b^2-a^2}}$
Now, $\cos ^{-1} x=\phi \Rightarrow x=\cos \phi$
$\Rightarrow x=\frac{b}{\sqrt{2 b^2-a^2}}$
$
\cot \left(\cos ^{-1} x\right)=\sec \left(\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}\right)
$
Let $\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}=\theta \Rightarrow \tan \theta=\frac{a}{\sqrt{b^2-a^2}}$
$\Rightarrow \sec \theta=\frac{b}{\sqrt{b^2-a^2}}$
$\therefore \quad \cot \left(\cos ^{-1} x\right)=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow \cos ^{-1} x=\cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)$
Again, let $\cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)=\phi \Rightarrow \cot \phi=\frac{b}{\sqrt{b^2-a^2}}$
$\Rightarrow \cos \phi=\frac{b}{\sqrt{2 b^2-a^2}}$
Now, $\cos ^{-1} x=\phi \Rightarrow x=\cos \phi$
$\Rightarrow x=\frac{b}{\sqrt{2 b^2-a^2}}$
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