Question:
If $\sin \theta+\cos \theta=\sqrt{2} \cos \left(90^{\circ}-\theta\right)$, find $\cot \theta$.
Solution:
Given: $\sin \theta+\cos \theta=\sqrt{2} \cos \left(90^{\circ}-\theta\right)$
We have to find the value of $\cot \theta$.
Now,
$\sin \theta+\cos \theta=\sqrt{2} \cos \left(90^{\circ}-\theta\right)$
$\Rightarrow \sin \theta+\cos \theta=\sqrt{2} \sin \theta$ (since, $\cos \left(90^{\circ}-\theta\right)=\sin \theta$ )
$\Rightarrow \cos \theta=(\sqrt{2}-1) \sin \theta$
$\Rightarrow \frac{\cos \theta}{\sin \theta}=\sqrt{2}-1$
$\Rightarrow \cot \theta=\sqrt{2}-1$
Hence, $\cot \theta=\sqrt{2}-1$