<s>[INST] Simplifying expression involving inverse trigonometric functions ${\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \left( {{{\tan }^{ - 1}}y} \right) + y\sin \left( {{{\tan }^{ - 1}}y} \right)}}{{\cot \left( {{{\sin }^{ - 1}}y} \right) + \tan \left( {{{\sin }^{ - 1}}y} \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} = \_\_\_\_\_$.
The official answer is $1$
My approach is as follow
${\tan ^{ - 1}}y = \gamma ;{\sin ^{ - 1}}y = \lambda  \Rightarrow \tan \gamma  = y\& \sin \lambda  = y$
$ \Rightarrow {\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \gamma  + y\sin \gamma }}{{\cot \lambda  + \tan \lambda }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \gamma  + \tan \gamma \sin \gamma }}{{\cot \lambda  + \tan \lambda }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {\frac{1}{{{{\tan }^2}\lambda }}{{\left( {\frac{1}{{\cos \gamma \left( {\cot \lambda  + \tan \lambda } \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\left( {\frac{1}{{\cos \gamma \left( {1 + {{\tan }^2}\lambda } \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {{{\left( {\frac{{{{\cos }^2}\lambda }}{{\cos \gamma }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {\frac{{{{\cos }^4}\lambda }}{{{{\cos }^2}\gamma }} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow \tan \gamma  = y\& \sin \lambda  = y$
$ \Rightarrow {\left( {\frac{{{{\cos }^4}\lambda }}{{{{\cos }^2}\gamma }} + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow \tan \gamma  = \sin \lambda  \Rightarrow {\tan ^2}\gamma  = {\sin ^2}\lambda  \Rightarrow {\sec ^2}\gamma  - 1 = {\sin ^2}\lambda  \Rightarrow \frac{1}{{{{\cos }^2}\gamma }} = 1 + {\sin ^2}\lambda $
$ \Rightarrow {\left( {\left( {1 + {{\sin }^2}\lambda } \right){{\cos }^4}\lambda  + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}} \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda  + {{\cos }^4}\lambda  + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda  - 2{{\sin }^2}\lambda {{\cos }^2}\lambda  + 2{{\sin }^2}\lambda {{\cos }^2}\lambda  + {{\cos }^4}\lambda  + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda  - 2{{\sin }^2}\lambda {{\cos }^2}\lambda  + 1} \right)^{\frac{1}{2}}}$
How do I proceed from here.
 [/INST] We can have the following reasoning as well:
If $\tan^{-1}y=A, -\dfrac\pi2<A<\dfrac\pi2$
$\tan(\tan^{-1}y)=\tan A=y$
$\cos(\tan^{-1}y)=+\dfrac1{\sqrt{\tan^2A+1}}=?$
$\sin(\tan^{-1}y)=?$
Similarly choose,  $\sin^{-1}y=B, -\dfrac\pi2\le B\le\dfrac\pi2$
Can you take it from here?
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