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Prove that:$ \left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\frac{1}{\sin ^{2} A-\sin ^{4} A} $
To do:
We have to prove that \( \left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\frac{1}{\sin ^{2} A-\sin ^{4} A} \).
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,
Let us consider LHS,
$\left(1+\tan ^{2} \mathrm{~A}\right)\left(1+\frac{1}{\tan ^{2} \mathrm{~A}}\right)=\left(1+\tan ^{2} \mathrm{~A}\right)+\left(1+\cot ^{2} \mathrm{~A}\right)$
$=\sec ^{2} \mathrm{~A} \operatorname{cosec}^{2} \mathrm{~A}$
Let us consider RHS,
$=\frac{1}{\sin ^{2} \mathrm{~A}-\sin ^{4} \mathrm{~A}}=\frac{1}{\sin ^{2} \mathrm{~A}-\sin ^{4} \mathrm{~A}}$
$=\frac{1}{\sin ^{2} A\left(1-\sin ^{2} A\right)}$
$=\frac{1}{\sin ^{2} A \cos ^{2} A}$
$=\operatorname{cosec}^{2} A \sec ^{2} A$
$=\sec ^{2} A \operatorname{cosec}^{2} A$
Here,
LHS $=$ RHS
Hence proved.
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