MCQ
$\frac { 1 + \tan ^ { 2 } A } { 1 + \cot ^ { 2 } A }=$
- A$\sec ^2 A$
- B$-1$
- C$\cot ^2 A$
- ✓$\tan ^2 A$
✓
Answer
Correct option: D.
$\tan ^2 A$
(d) $\tan ^2 A$
$\begin{array}{l}\frac{1+\tan ^2 A}{1+\cot ^2 A}=\frac{1+\frac{\sin ^2 A}{\cos ^2 A}}{1+\frac{\cos ^2 A}{\sin ^2 A}} \\
=\frac{\frac{\cos ^2 A+\sin ^2 A}{\cos ^2 A}}{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A}} \\
=\frac{\frac{1}{\cos ^2 A}}{\frac{1}{\sin ^2 A}} \\
=\frac{\sin ^2 A}{\cos ^2 A} \\
=\tan ^2 A\end{array}$
Hence, alternative $\tan ^2 A$ is correct.
$\begin{array}{l}\frac{1+\tan ^2 A}{1+\cot ^2 A}=\frac{1+\frac{\sin ^2 A}{\cos ^2 A}}{1+\frac{\cos ^2 A}{\sin ^2 A}} \\
=\frac{\frac{\cos ^2 A+\sin ^2 A}{\cos ^2 A}}{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A}} \\
=\frac{\frac{1}{\cos ^2 A}}{\frac{1}{\sin ^2 A}} \\
=\frac{\sin ^2 A}{\cos ^2 A} \\
=\tan ^2 A\end{array}$
Hence, alternative $\tan ^2 A$ is correct.
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