Question

Prove that $cosec\left({\tan }^{-1}\right(\cos \left({\cot }^{-1}\right(\sec \left({\sin }^{-1}a\right)\left)\right)\left)\right)=\sqrt{3-a^2}$, where $a\in [0,1]$.

Answer 1

given, $cosec\left({\tan }^{-1}\right(\cos \left({\cot }^{-1}\right(\sec \left({\sin }^{-1}a\right)\left)\right)\left)\right)$

$=cosec\left({\tan }^{-1}\left(\cos \left({\cot }^{-1}\left(\sec \left({\sec }^{-1}\frac{1}{\sqrt{1-a^2}}\right)\right)\right)\right)\right)$ {Since for $a\in [0,1],{\sin }^{-1}a={\sec }^{-1}\frac{1}{\sqrt{1-a^2}}$]

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

$=cosec\left({\tan }^{-1}\left(\cos \left({\cos }^{-1}\frac{1}{\sqrt{2-a^2}}\right)\right)\right)$ [Since for $a\in [0,1],{\cot }^{-1}\frac{1}{\sqrt{1-a^2}}={\cos }^{-1}\frac{1}{\sqrt{2-a^2}}$]

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

$=cosec\left(cose{c}^{-1}\sqrt{3-a^2}\right)$ [Since for $a\in [0,1],{\tan }^{-1}\frac{1}{\sqrt{2-a^2}}=cose{c}^{-1}\sqrt{3-a^2}$]

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