Question
Prove that $( 1 + \tan A)^2 + (1 - \tan A)^2 = 2 \sec^2A$
✓
Answer
$LHS = ( 1 + \tan A)^2 + (1 - \tan A)^2$
$= 1 + 2 \tan A + \tan^2A + 1 - 2 \tan A + \tan^2A$
$= 2( 1 + \tan^2A)$
$= 2 \sec^2A$
$= RHS$
Hence proved.
$= 1 + 2 \tan A + \tan^2A + 1 - 2 \tan A + \tan^2A$
$= 2( 1 + \tan^2A)$
$= 2 \sec^2A$
$= RHS$
Hence proved.
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