Question

tan^2A.sec^2B -sec^2Atan^2B =tan^2A - tan^2B

Answer 1

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Answer 2

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

Step-by-step explanation:

To prove that, $=\tan^2A \sec^2B -\sec^2A \tan^2B=\tan^2A -\tan^2B$

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

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

[ ∵ $\sec^{2} \theta- \tan^{2} \theta=1$

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

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

=L.H.S.

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