Question

$x=co\sec \left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\sec \left({\sin }^{-1}a\right)\right)\right)\right)\right);y=\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(co\sec \left({\cos }^{-1}a\right)\right)\right)\right)\right)$, where $a\in [0,1]$. Then which of the following options is correct?

• A. $x^2+a^2=3$
• B. $x=y$
• C. $y^2+a^2=3$
• D. All of these

Answer 1

The correct option is D All of these

$x=\csc \left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\sec \left(\frac{{\sin }^{-1}a}{\theta }\right)\right)\right)\right)\right)$

$\sin \theta =a$

$x=\csc \left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)$

$\cot \alpha =\frac{1}{\sqrt{1-a^2}}$

$x=\csc \frac{\left({\tan }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)}{\beta }$

$\tan \beta =\frac{1}{\sqrt{2-a^2}}$

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

$x^2+a^2=3$

$y=\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(\csc \left(\frac{{\cos }^{-1}a}{\theta }\right)\right)\right)\right)\right)$

$\cos \theta =a$

$y=\frac{\sec \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}\left(\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)}{\alpha }$

$\tan \alpha =\frac{1}{\sqrt{1-a^2}}$

$y=\frac{\sec \left({\cot }^{-1}\left(\frac{1}{\sqrt{2-a^2}}\right)\right)}{\beta }$

$\cot \beta =\frac{1}{\sqrt{2-a^2}}$

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

$y^2+a^2=3$

$x=y$