Question

prove that tan2A-tan2B=(sin2A-sin2B)/(cos2A*cos2B)

Answer 1

i hope it helps. Sorry for the strikes

Answer 2

Answer with Step-by-step explanation:

We have to prove that

$tan^2A-tan^2B=\frac{sin^2A-sin^2B}{cos^2Acos^2B}$

Substitute $tan\theta=\frac{sin\theta}{cos\theta}$

LHS

=$\frac{sin^2A}{cos^2B}-\frac{sin^2B}{cos^2A}$

=$\frac{sin^2cos^2B-sin^2Bcos^2A}{cos^2Acos^2B}$

=$\frac{sin^2A(1-sin^2B)-(1-sin^2A)(sin^2B}{cos^2Acos^2B}$

Using identity :$cos^2A=1-sin^2A$

=$\frac{sin^2A-sin^2Asin^2B-sin^2B+sin^2Asin^2B}{cos^2Acos^2B}$

=$\frac{sin^2A-sin^2B}{cos^2Acos^2B}$

Hence, proved.