Question

how to prove this question.. we have to use trignometrical identities

Answer 1

at First convert tan square A into sin square A by cos square A.

$\frac{{sin}^{2} a - ( {sina}^{2} \times {cosa}^{2} )}{ {cos}^{2} a}$



[sin^2 a (1-cos^2 a)]/cos^2 a


we know that 1-cos^2 a = sin^2a
so
$\frac{ {sin}^{2}a \times {sin}^{2} a }{ {cos}^{2}a } = \frac{ {sin}^{2} a}{ {cos}^{2} a} \times {sin}^{2} a$
hence
${tan}^{2} a. {sin}^{2} a$
hope you got it