Prove the following trigonometric identities.
$\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1$
We have to prove $\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=1$
We know that, $\sin ^{2} A+\cos ^{2} A=1$
So,
$\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A=\frac{1}{\cos ^{4} A}\left(1-\sin ^{4} A\right)-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\left(\frac{1}{\cos ^{4} A}-\frac{\sin ^{4} A}{\cos ^{4} A}\right)-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\left(\frac{1-\sin ^{4} A}{\cos ^{4} A}\right)-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\left(1-\sin ^{2} A\right)\left(1+\sin ^{2} A\right)}{\cos ^{4} A}-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A\left(1+\sin ^{2} A\right)}{\cos ^{4} A}-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1+\sin ^{2} A}{\cos ^{2} A}-2 \frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1+\sin ^{2} A-2 \sin ^{2} A}{\cos ^{2} A}$
$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}$
$=1$
Hence proved.
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