Question

Prove that
$\cos (\frac{3\pi }{2}+\theta )\cos (2\pi +\theta )[\cot (\frac{3\pi }{2}-\theta )+\cot (2\pi +\theta \left)\right]=1$

Answer 1

L.H.S

$\Rightarrow \cos (\frac{3\pi }{2}+\theta )\cos (2\pi +\theta )[\cot (\frac{3\pi }{2}-\theta )+\cot (2\pi +\theta )]$

We know that

$\cos (\frac{3\pi }{2}+\theta )=\sin \theta$

$\cos (2\pi +\theta )=\cos \theta$

$\cot (\frac{3\pi }{2}-\theta )=\tan \theta$

$\cot (2\pi +\theta )=\cot \theta$

Therefore,

$\Rightarrow \sin \theta \cos \theta (\tan \theta +\cot \theta )$

$\Rightarrow \sin \theta \cos \theta (\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })$

$\Rightarrow \sin \theta \cos \theta \left(\frac{{\sin }^2\theta +{\cos }^2\theta }{sin\theta \cos \theta }\right)$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta$

$\Rightarrow 1$

Hence, this is the answer.