Prove the following : (1 + tanA tanB)^2 + (tanA – tanB)^2 = ^2A ^… - Vidyadip

Question

Prove the following : $(1 + tanA tanB)^2 + (tanA – tanB)^2 = \sec ^2A \sec^2B$

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Answer

L.H.S. $= (1 + tanA tanB)^2 + (tanA – tanB)^2$
$= 1 + 2tanA tanB + \tan^2A \tan^2 + \tan^2 A- 2tanA tanB + \tan^2B$
$= 1 + \tan^2A + \tan^2 B + \tan^2A \tan^2B$
$= 1(1+ \tan^2A) + \tan^2 B(1 + \tan^2A)$
$= (1 + \tan^2A) (1 + \tan^2B)$
$= \sec^2A \sec^2B$ = R.H.S.

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