If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =

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 $(b)$