Question

show that [1+tan ^2 A / 1+cot^2]
= [1+tanA/1+cotA]^2= tan^2​

Answer 1

$\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A} \\ \\ = \frac{ {sec}^{2} A}{ {cosec}^{2}A } \\ \\ = \frac{ {sin}^{2} A}{ {cos}^{2}A } \\ \\ = {tan}^{2} A \\ \\ \\ ( { \frac{1 + tanA}{1 + cotA} })^{2} \\ \\ = ( { \frac{ \frac{cosA + sinA}{cosA} }{ \frac{sinA + cosA}{sinA} } })^{2} \\ \\ = ( { \frac{sinA}{cosA} })^{2} \\ \\ = {tan}^{2} A$