Prove that (cosec θ − sin θ) (sec θ − cos θ) =
Question:

Prove that $(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)=\frac{1}{\tan \theta+\cot \theta}$.

Solution:

Here we have to prove that

$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)=\frac{1}{\tan \theta+\cot \theta}$

Left hand side

$=(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)$

$=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)$

$=\frac{\left(1-\sin ^{2} \theta\right)\left(1-\cos ^{2} \theta\right)}{\sin \theta \cos \theta}$

Now using the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

Left hand side

$=\frac{\sin ^{2} \theta \cos ^{2} \theta}{\sin \theta \cos \theta}$

$=\frac{\sin \theta \cos \theta}{1}$

$=\frac{\sin \theta \cos \theta}{\sin ^{2} \theta+\cos ^{2} \theta}$

$=\frac{1}{\frac{\sin ^{2} \theta}{\sin \theta \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cos \theta}}$

$=\frac{1}{\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}}$

$=\frac{1}{\tan \theta+\cot \theta}$

Hence proved.

 

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