Prove the following identities:$ \sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1 $

To do:

We have to prove that \( \sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1 \)

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,

$\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}=\frac{1}{\cos ^{4} \mathrm{~A}}\left(1-\sin ^{4} \mathrm{~A}\right)- \frac{2\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{1}{\cos ^{4} \mathrm{~A}}\left(1-\sin ^{2} \mathrm{~A}\right)\left(1+\sin ^{2} \mathrm{~A}\right)-\frac{2\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{1}{\cos ^{4} \mathrm{~A}}\left(\cos ^{2} \mathrm{~A}\right)\left(1+\sin ^{2} \mathrm{~A}\right)-\frac{2\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{1+\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}-\frac{2\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{1+\sin ^{2} \mathrm{~A}-2 \sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$

$=1$

 Hence proved.