Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
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Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
 A: Then we have $$\cos\left(\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)\right)$$
If $\tan^{-1}\dfrac1{\sqrt{1+x^2}}=y,\dfrac\pi4\le y\le\dfrac\pi2\implies\tan y=\dfrac1{\sqrt{1+x^2}}$ and $\cos y\ge0$
$\cos\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)=\cos y=+\dfrac1{\sqrt{1+\tan^2y}}=?$
A: Let $y=\frac{1}{\sqrt{1+x^2}}$ and let $\tan^{-1}y=w$
Then, $$\begin{align}
\ \cos{\tan^{-1}y} & =\cos{w} \\
\ \text{Now since $1 \ge y \gt 0$ we have $\cos{w} \gt 0$}\\
\ &= \frac{1}{\sec{w}}\\
\ &= \frac{1}{\sqrt{1+\tan^2{w}}}\\
\ &= \frac{1}{\sqrt{1+y^2}}\
\end{align}$$
And then substitute $y$.
A: So your final expression is: $\cos\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)$
Now we know that: 
$\tan^{-1}p=\cos^{-1}\left(\dfrac{1}{\sqrt{1+p^2}}\right)\\
\implies\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+{\left(\dfrac{1}{\sqrt{1+x^2}}\right)}^2}}\right)=\cos^{-1}\left(\dfrac{1}{\sqrt{1+{\dfrac{1}{1+x^2}}}}\right)\\
\implies\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos^{-1}\sqrt{\dfrac{1+x^2}{2+x^2}}$
Therefore, $\cos\tan^{-1}\left(\dfrac{1}{\sqrt{1+x^2}}\right)=\cos\cos^{-1}\sqrt{\dfrac{1+x^2}{2+x^2}}=\sqrt{\dfrac{1+x^2}{2+x^2}}$
