Question

prove that sin7x+sin5x+sin9x+sin3x/cos7x+cos5x+cos9x+cos3x=tan6x​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\:\dfrac{sin7x + sin5x + sin9x + sin3x}{cos7x + cos5x +cos9x + cos3x}$

$\rm \: = \:\dfrac{(sin7x + sin5x) + (sin9x + sin3x)}{(cos7x + cos5x) +(cos9x + cos3x)}$

We know,

$\boxed{ \sf{ \:cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$

$\boxed{ \sf{ \:sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$

Using these results, we get

$\rm \:=\dfrac{2sin\bigg(\dfrac{7x + 5x}{2} \bigg)cos\bigg(\dfrac{7x - 5x}{2} \bigg) + 2sin\bigg(\dfrac{9x + 3x}{2} \bigg)cos\bigg(\dfrac{9x - 3x}{2} \bigg)}{2cos\bigg(\dfrac{7x + 5x}{2} \bigg)cos\bigg(\dfrac{7x - 5x}{2} \bigg) + 2cos\bigg(\dfrac{9x + 3x}{2} \bigg)cos\bigg(\dfrac{9x - 3x}{2} \bigg)}$

$\rm \:  =  \:  \:\dfrac{2sin6xcosx + 2sin6xcos3x}{2cos6xcosx + 2cos6xcos3x}$

$\rm \:  =  \:  \:\dfrac{2sin6x(cosx + cos3x)}{2cos6x(cosx + cos3x)}$

$\rm \:  =  \:  \:\dfrac{sin6x}{cos6x}$

$\rm \:  =  \:  \:tan6x$

Hence, Proved