Question

please answer my question

Answer 1

hope this is right ☺☺☺

Answer 2

hey

here is answer.

to prove -

${ \tan}^{2} a - {tan}^{2} b = \frac{ {sin}^{2}a - {sin}^{2}b }{cos ^{2} a- cos ^{2} b}$


we know that.

tan= sin/cos

lhs

$(\frac{sina}{cosa})^{2} - (\frac{sinb}{cosb})^{2}$

$\frac{sin ^{2} a - sin \: ^{2}b }{cos ^{2}a \times {cos}^{2} b }$

hope it helps


thanks