MCQ
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$ is equal to:
- A$\sec^2\text{A}$
- B$-1$
- C$\cot^2\text{A}$
- ✓$\tan^2\text{A}$
✓
Answer
Correct option: D.
$\tan^2\text{A}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}$
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}$
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
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