Prove that:$ (\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\sec \theta+\operatorname{cosec} \theta $

To do:

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

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

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

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

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

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

Therefore,

$(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

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

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

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

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

$= \sec \theta+\cos \theta$

Hence proved.  

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Updated on: 10-Oct-2022

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