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