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.

• you made a mistake in line $2$, $y = \tan \gamma$ but not $\tan \lambda$ – Learning Mathematics Oct 25 '20 at 15:32
• I understood my mistake let me try it again – Samar Imam Zaidi Oct 25 '20 at 16:33
• @Learning Mathematics Thanks for pointing my mistake. Now getting the answer 1. – Samar Imam Zaidi Oct 27 '20 at 3:18

We can have the following reasoning as well:

If $$\tan^{-1}y=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?