Prove the following:
$ \frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}-\frac{\tan \mathrm{A}}{1-\sec \mathrm{A}}=2 \operatorname{cosec} \mathrm{A} $

To do:

We have to prove that \( \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \operatorname{cosec} A \).

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,

$\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}-\frac{\tan \mathrm{A}}{1-\sec \mathrm{A}}=\frac{\tan \mathrm{A}(1-\sec \mathrm{A}-1-\sec \mathrm{A})}{(1+\sec \mathrm{A})(1-\sec \mathrm{A})}$

$=\frac{\tan \mathrm{A}(-2 \sec \mathrm{A})}{\left(1-\sec ^{2} \mathrm{~A}\right)}$

$=\frac{2 \tan \mathrm{A}\sec \mathrm{A}}{\left(\sec ^{2} \mathrm{~A}-1\right)}$

$=\frac{2 \tan \mathrm{A} \cdot \sec \mathrm{A}}{\tan ^{2} \mathrm{~A}}$

$=\frac{2 \sec \mathrm{A}}{\tan \mathrm{A}}$

$=2\times\frac{1}{\cos A}\times\frac{\cos A}{\sin A}$

$=\frac{2}{\sin \mathrm{A}}$

$=2 \operatorname{cosec} \mathrm{A}$

Hence proved.    

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Updated on: 10-Oct-2022

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