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:

You can also solve this by converting \(\cot^2 A\) into its reciprocal form, \(\frac{1}{\tan^2 A}\):
\[ \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}} = \frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \times \frac{\tan^2 A}{1 + \tan^2 A} = \tan^2 A \] This method requires fewer identity changes and solves the expression in just two steps!

Answer 1

The Correct Option is A

Step 1: Rewrite tan and cot in terms of sin and cos.
\[ 1 + \tan^2 A = 1 + \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A + \sin^2 A}{\cos^2 A} = \frac{1}{\cos^2 A} \]
Step 2: Do the same for the denominator.
\[ 1 + \cot^2 A = 1 + \frac{\cos^2 A}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A} \]
Step 3: Divide the two results.
\[ \frac{1+\tan^2 A}{1+\cot^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} \]
Step 4: Simplify to the final form.
\[ \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \]
This matches option (A).
\[ \boxed{\tan^2 A} \]