If $x \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)=\cos \left(90^{\circ}-\theta\right)$, then $x=$
(a) 0
(b) 1
(c) $-1$
(d) 2
We have: $x \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)=\cos \left(90^{\circ}-\theta\right)$
Here we have to find the value of $x$
We know that $\left[\begin{array}{l}\sin \left(90^{\circ}-\theta\right)=\cos \theta \\ \cos \left(90^{\circ}-\theta\right)=\sin \theta \\ \cot \left(90^{\circ}-\theta\right)=\tan \theta\end{array}\right]$
$\Rightarrow x \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)=\cos \left(90^{\circ}-\theta\right)$
$\Rightarrow x \cos \theta \tan \theta=\sin \theta$
$\Rightarrow x \cos \theta \times \frac{\sin \theta}{\cos \theta}=\sin \theta$
$\Rightarrow x=1$
Hence the correct option is $(b)$