Prove the following identities:$ (\sec A+\tan A-1)(\sec A-\tan A+1)=2 \tan A $

To do:

We have to prove that \( (\sec A+\tan A-1)(\sec A-\tan A+1)=2 \tan A \).

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,

$(\sec A+\tan A-1)(\sec A-\tan A+1) =[(\sec A)+(\tan A-1)][(\sec A)-(\tan A-1)]$

$=(\sec A)^{2}-(\tan A-1)^{2}$

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

$=\sec ^{2} A-\tan ^{2} A-1+2 \tan A$

$=1-1+2 \tan A$

$=2 \tan A$

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Updated on: 10-Oct-2022

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