Question

\(\frac{1+tan^2A}{1+cot^2A}\) =

• A.
• B. -1
• C.
• D.

Answer 1

The correct option is D

$\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}$

=$\frac{1+ta{n}^{2}A}{1+\frac{1}{ta{n}^{2}A}}$

=$\frac{1+ta{n}^{2}A}{\frac{ta{n}^{2}A+1}{ta{n}^{2}A}}$

=$\frac{(1+ta{n}^{2}A)\left(ta{n}^{2}A\right)}{(ta{n}^{2}A+1)}$

= $ta{n}^{2}A.$