MCQ
$\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}$ बराबर है:
  • A
    $-1$
  • B
    $\sec ^2 A$
  • $\tan ^2 A$
  • D
    $\cot ^2 A$

Answer

Correct option: C.
$\tan ^2 A$
$\frac{1+\tan ^2 A}{1+\cot ^2 A}=\frac{\sec ^2 A}{\operatorname{cosec}^2 A}$
$=\frac{\frac{1}{\cos ^2 A}}{\frac{1}{\sin ^2 A}}=\frac{1}{\cos ^2 A} \times \frac{\sin ^2 A}{1}$
$=\frac{\sin ^2 A}{\cos ^2 A}=\tan ^2 A$

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