Question

If $x=\text{cosec}\left({\tan }^{-1}\right(\cos \left({\cot }^{-1}\right(\sec \left({\sin }^{-1}a\right)\left)\right)\left)\right)$ and $y=\sec \left({\cot }^{-1}\right(\sin \left({\tan }^{-1}\right(\text{cosec}\left({\cos }^{-1}a\right)\left)\right)\left)\right),$ where $a\in [0,1],$ then which of the following options is(are) CORRECT?

• A. $x=\sqrt{2-a^2}$
• B. $x=\sqrt{3-a^2}$
• C. $y=\frac{1}{\sqrt{2-a^2}}$
• D. $y=\sqrt{3-a^2}$

Answer 1

Answer: (B), (D)

$y=\sqrt{3-a^2}$

$x=\text{cosec}\left({\tan }^{-1}\right(\cos \left({\cot }^{-1}\right(\sec \left({\sin }^{-1}a\right)\left)\right)\left)\right)$
$\Rightarrow x=\text{cosec}\left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\sec \left({\sec }^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)\right)\right)$
$\Rightarrow x=\text{cosec}\left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)$
$\Rightarrow x=\text{cosec}\left({\tan }^{-1}\left(\cos \left({\cos }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)\right)\right)$
$\Rightarrow x=\text{cosec}\left({\tan }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)$
$\Rightarrow x=\text{cosec}\left({\text{cosec}}^{-1}\left(\sqrt{3-a^2}\right)\right)$
$\Rightarrow x=\sqrt{3-a^2}$

$y=\sec \left({\cot }^{-1}\right(\sin \left({\tan }^{-1}\right(\text{cosec}\left({\cos }^{-1}a\right)\left)\right)\left)\right)$
$\Rightarrow y=\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(\text{cosec}\left({\text{cosec}}^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)\right)\right)$
$\Rightarrow y=\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)$
$\Rightarrow y=\sec \left({\cot }^{-1}\left(\sin \left({\sin }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)\right)\right)$
$\Rightarrow y=\sec \left({\cot }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)$
$\Rightarrow y=\sec \left({\sec }^{-1}\left(\sqrt{3-a^2}\right)\right)$
$\Rightarrow y=\sqrt{3-a^2}$