# Prove the following:$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$

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Given:

$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$

To do:

We have to prove that $(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$.

Solution:

We know that,

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

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

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

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

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

LHS $=(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)$

$=(\sin \alpha+\cos \alpha)(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha})$

$=(\sin \alpha+\cos \alpha)(\frac{\sin^2 \alpha+\cos^2 \alpha}{\sin \alpha \cos \alpha})$

$=(\sin \alpha+\cos \alpha) \frac{1}{(\sin \alpha \cos \alpha)}$

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

$=\sec \alpha+\operatorname{cosec} \alpha$

$=$ RHS

Hence proved.

Updated on 10-Oct-2022 13:29:22