Question

If $\sin \left[{\cot }^{-1}\right(x+1\left)\right]=\cos \left({\tan }^{-1}x\right)$, then find $x$.

Answer 1

$sin\left[co{t}^{-1}\right[1+x\left]\right]=cos\left[ta{n}^{-1}x\right]$

Now

$co{t}^1(1+x)=p$ & $ta{n}^{-1}x=9$

$cotp=1+x$ tanq=x

$sinp=\frac{1}{\sqrt{(1+x{)}^2+1}}$ $sinq=\frac{x}{\sqrt{1+x^2}};cosq=\frac{1}{\sqrt{1+x^2}}$

then

$sin\left[si{n}^{-1}\frac{1}{\sqrt{(1+x{)}^2+1}}\right]=cos\left[co{s}^{-1}\frac{1}{\sqrt{1+x^2}}\right]$

$\Rightarrow \frac{1}{\sqrt{(1+x{)}^2+1}}=\frac{1}{\sqrt{1+x^1}}$

squaring both side we get

$\Rightarrow 1+x^2=(1+x{)}^2+x$

$x^2=1+x^2+2x$

2x=-1

$[x=\frac{-1}{2}]$