Question

Prove that{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)

Answer 1

First, expand the terms inside the bracket you will get

$(( \tan {}^{2} (x) + 2 \tan(x) \sin(x) + \sin {}^{2} (x) - ( \tan {}^{2} (x) - 2 \tan(x) + \sin {}^{2} (x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$( 4 \tan(x) \sin(x) ) {}^{2} = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) \sin {}^{2} (x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16 \tan {}^{2} (x) (1 - \cos {}^{2} (x) ) = 16 (\tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan {}^{2} (x) - \frac{ \sin {}^{2} (x) \cos {}^{2} ( {x}^{} ) }{ \cos {}^{2} (x) }$

$16( \tan {}^{2} (x) - \sin {}^{2} (x) ) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$

$16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) = 16( \tan(x) + \sin(x) )( \tan(x) - \sin(x) )$