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) )`