MCQ
$\frac{1+\cot ^2 A }{1+\tan ^2 A }=?$
- A$\tan ^2 \mathring A$
- B$\sec ^2 \mathring A$
- C$\operatorname{cosec}^2 \mathring A$
- ✓$\cot ^2 \mathring A$
✓
Answer
Correct option: D.
$\cot ^2 \mathring A$
$\cot ^2 \mathrm{~A}$$\frac{1+\cot ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}$
$=\frac{\operatorname{cosec}^2 \mathrm{~A}}{\sec ^2 \mathrm{~A}}$
$=\frac{\frac{1}{\sin ^2 \mathrm{~A}}}{\frac{1}{\cos ^2 \mathrm{~A}}}$
$=\frac{\cos ^2 \mathrm{~A}}{\sin ^2 \mathrm{~A}}$
$=\cot ^2 \mathrm{~A}$
$=\frac{\operatorname{cosec}^2 \mathrm{~A}}{\sec ^2 \mathrm{~A}}$
$=\frac{\frac{1}{\sin ^2 \mathrm{~A}}}{\frac{1}{\cos ^2 \mathrm{~A}}}$
$=\frac{\cos ^2 \mathrm{~A}}{\sin ^2 \mathrm{~A}}$
$=\cot ^2 \mathrm{~A}$
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