Question

\(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) equals to :

• A. \(\tan^2 A\)
• B. –1
• C. \(-\tan^2 A\)
• D. \(\cot^2 A\)

Hint:

An alternate quick way to solve this is to write \(\cot^2 A\) as the reciprocal of \(\tan^2 A\):
\[ \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}} = \frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \cdot \frac{\tan^2 A}{1 + \tan^2 A} = \tan^2 A \] This method requires only one trigonometric identity change and is extremely elegant!

Answer 1

The Correct Option is A

Step 1: Write cotangent as the reciprocal of tangent.
$\cot^2A=\frac{1}{\tan^2A}$, so $1+\cot^2A=\frac{\tan^2A+1}{\tan^2A}$.
Step 2: Rewrite the whole expression as one division.
$\frac{1+\tan^2A}{1+\cot^2A}=(1+\tan^2A)\times\frac{\tan^2A}{1+\tan^2A}$.
Step 3: Cancel the common factor.
The $(1+\tan^2A)$ terms cancel, leaving $\tan^2A$, matching option (A).
\[ \boxed{\tan^2 A} \]