Answer:
See Explanation
Step-by-step explanation:
[tex]
\because \sin\bigg(\frac{\pi}{2} + x\bigg)= \cos x\\
\&\: \sin (\pi - x) = \sin x\\\\
Now,
\frac{\sin\bigg(\frac{\pi}{2} + x\bigg)}{\sin (\pi - x)} = \cot x\\\\
LHS
= \frac{\sin\bigg(\frac{\pi}{2} + x\bigg)}{\sin (\pi - x)} \\\\
=\frac{\cos x}{sinx} \\\\
= \cot x\\\\
= RHS\\\\
Hence\: Proved
[/tex]
