Prove that:$ (\sec A-\tan A)^{2}=\frac{1-\sin A}{1+\sin A} $

To do:

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

Solution:

We know that,

$\sin ^{2} A+\cos^2 A=1$.......(i)

$\sec A=\frac{1}{\cos A}$......(ii)

$\tan A=\frac{\sin A}{\cos A}$......(iii)

Therefore,

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

$=\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^{2}$

$=\left(\frac{1-\sin A}{\cos A}\right)^{2}$

$=\frac{(1-\sin A)^{2}}{\cos ^{2} A}$

$=\frac{(1-\sin A)^{2}}{1-\sin ^{2} A}$

$=\frac{(1-\sin A)(1-\sin A)}{(1+\sin A)(1-\sin A)}$

$=\frac{1-\sin A}{1+\sin A}$

Hence proved.     

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Updated on: 10-Oct-2022

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