Question

prove that cos4x + cos3x + cos2x /
sin4x + sin3x + sin2x = cot3x​

Answer 1

(cos4x+cos3x+cos2x)/(sin4x+sin3x+sin2x)=cotx

LHS

=(cos4x+cos2x+cos3x)/(sin4x+sin2x+sin3x)

=[2cos3x.cosx+cos3x]/[2sin3x.cosx+sin3x]

=[cos3x(2cosx+1)]/[sin3x(2cosx+1)]

=cos3x/sin3x

=cot3x , proved

MARK ME AS BRAINLIEST PLEASE

Answer 2

Question

Prove that $\sf \dfrac{cos4x + cos3x + cos2x}{ sin4x + sin3x + sin2x} = cot3x$

Answer

We know,

$\sf sinC + sinD = 2sin \bigg( \dfrac{C + D}{2} \bigg) cos \bigg(\dfrac{C - D}{2} \bigg)$

$\sf cosC + cosD = 2cos\bigg( \dfrac{C + D}{2} \bigg) cos \bigg(\dfrac{C - D}{2} \bigg)$

Given,

$\sf \dfrac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x}$

$\sf = \dfrac{(cos4x + cos2x )+ cos3x}{(sin4x + sin2x) + sin3x}$

$\sf = \dfrac{2cos \bigg( \dfrac{4x + 2x}{2} \bigg)cos \bigg( \dfrac{4x - 2x}{2} \bigg) cos3x }{ 2sin \bigg( \dfrac{4x + 2x}{2} \bigg)cos \bigg( \dfrac{4x - 2x}{2} \bigg) sin3x }$

$\sf = \dfrac{2cos \bigg( \dfrac{6x}{2} \bigg)cos \bigg( \dfrac{2x}{2} \bigg) cos3x }{ 2sin \bigg( \dfrac{6x}{2} \bigg)cos \bigg( \dfrac{2x}{2} \bigg) sin3x }$

$\sf = \dfrac{2cos3x \: cosx + cos3x }{ 2sin 3x \: cosx + sin3x }$

$\sf = \dfrac{cos3x (2 cosx + 1)}{ sin 3x (2 cosx + 1) }$

$\sf = \dfrac{cos3x}{sin 3x }$

$\sf = cot3x$

Hence proved !