Question

tan2Asec2B-sec2Atan2B=tan2A-tan2B If you solve it I will mark you as brainlist definitely.

Answer 1

That will help you reply dena sahi hai ki nahi

Answer 2

$\tan^2A\sec^2B-\sec^2A\tan^2B=\tan^2A-\tan^2B,$ proved.

Step-by-step explanation:

Prove that, $\tan^2A\sec^2B-\sec^2A\tan^2B=\tan^2A-\tan^2B$.

L.H.S.$=\tan^2A\sec^2B-\sec^2A\tan^2B$

$=\tan^2A(1+\tan^2B)-(1+\tan^2A)\tan^2B$

Using trigonometric identuty,

$\sec^2 \theta-\tan^2 \theta=1$

$=\tan^2A+\tan^2A\tan^2B-(\tan^2B+\tan^2A\tan^2B)$

$=\tan^2A+\tan^2A\tan^2B-\tan^2B-\tan^2A\tan^2B$

$=\tan^2A-\tan^2B$

=R.H.S., proved.

Hence, $\tan^2A\sec^2B-\sec^2A\tan^2B=\tan^2A-\tan^2B$, proved.