Question
Prove that $\frac{1+\tan ^2 A}{1+\cot ^2 A}=\sec ^2 A-1$
✓
Answer
We know, trigonometric identities
$
1+\tan ^2 A=\sec ^2 A \text { and } 1+\cot ^2 A=\operatorname{cosec}^2 A
$
Therefore, the given L.H.S. becomes
$
\begin{aligned}
\frac{\sec ^2 A}{\operatorname{cosec}^2 A} & =\frac{\sin ^2 A}{\cos ^2 A} \\
& =\tan ^2 A \\
& =1-\sec ^2 A
\end{aligned}
$
Hence proved.
$
1+\tan ^2 A=\sec ^2 A \text { and } 1+\cot ^2 A=\operatorname{cosec}^2 A
$
Therefore, the given L.H.S. becomes
$
\begin{aligned}
\frac{\sec ^2 A}{\operatorname{cosec}^2 A} & =\frac{\sin ^2 A}{\cos ^2 A} \\
& =\tan ^2 A \\
& =1-\sec ^2 A
\end{aligned}
$
Hence proved.
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