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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.
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