Prove. (sec A - A) (sec A + A) = ^2 A + ^2 A - Vidyadip

Question

Prove. $(sec A - \cos A) (sec A + \cos A) = \sin^2 A + \tan^2 A$

✓

Answer

$ \text{LHS} =(sec A - \cos A) (sec A + \cos A)$
$= sec^2 A - \cos^2 A$
$= (1 + \tan^2 A) - (1 - \sin^2 A)$
$= \sin^2 A + \tan^2 A = RHS$

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