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: Convert everything to sine and cosine right away.
$\tan^2 A = \frac{\sin^2 A}{\cos^2 A}$ and $\cot^2 A = \frac{\cos^2 A}{\sin^2 A}$.
Step 2: Simplify the numerator and denominator separately.
Numerator: $1 + \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A} = \frac{1}{\cos^2 A}$.
Denominator: $1 + \frac{\cos^2 A}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A}$.
Step 3: Divide the two results.
\[ \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \]
\[ \boxed{\tan^2 A} \]