Question

Prove that $\sin {\cot }^{-1}\tan {\cos }^{-1}x=\sin cose{c}^{-1}\cot {\tan }^{-1}x=x$ where $xϵ(0,1]$

Answer 1

Let $x=cos\theta$. where $cos\theta ϵ(0,1]$
Hence $\theta ϵ[0,\frac{\pi }{2})\cup (\frac{3\pi }{2},2\pi ]$.
Then
$sin\left(co{t}^{-1}\right(tan\left(co{s}^{-1}\right(x\left)\right)\left)\right)$
$=sin\left(co{t}^{-1}\right(tan\left(co{s}^{-1}\right(cos\theta \left)\right)\left)\right)$
$=sin\left(co{t}^{-1}\right(tan\theta \left)\right)$
$=sin(\frac{\pi }{2}-ta{n}^{-1}(tan\theta \left)\right)$
$=sin(\frac{\pi }{2}-\theta )$
$=cos\theta =x$ ...(i)
Similarly
$sin\left(cose{c}^{-1}\right(cot\left(ta{n}^{-1}x\right)\left)\right)$
Let $x=tan\theta$. where $tan\theta ϵ(0,1]$
Hence $\theta ϵ(0,\frac{\pi }{4}]\cup (\pi ,\frac{5\pi }{4}]$
Then
$sin\left(cose{c}^{-1}\right(cot\left(ta{n}^{-1}x\right)\left)\right)$
$=sin\left(cose{c}^{-1}\right(cot\left(ta{n}^{-1}tan\theta \right)\left)\right)$
$=sin\left(cose{c}^{-1}\right(cot\theta \left)\right)$
$=\frac{1}{cosec\left(cose{c}^{-1}\right(cot\theta \left)\right)}$
$=\frac{1}{cot\theta }$
$=tan\theta$
$=x$.