Question

cos4x+cos3x+cos2x/sin4x+sin3x+sin2x=cot3x

Answer 1

hope this helps you☺️☺️

Answer 2

$\bold{\huge{\underline{Solution-}}}$

LHS=

$\displaystyle \sf \: { \frac{cos4x + cos3x + cos2x}{sin4x + sin3x + sin2x} }$

$= \displaystyle \sf \: { \frac{cos3x + \left( \: cos4x + cos2x \right)}{sin3x + \left( sin4x + sin2x\right)} } \\ \\ = \displaystyle \sf \frac{cos3x + 2cos \left( \frac{4x + 2x}{2} \right)cos \left( \frac{4x - 2x}{2} \right)}{sin3x + 2sin \left( \frac{4x + 2x}{2} \right)cos \left( \frac{4x - 2x}{2} \right)} \\ \\ = \displaystyle \sf \: \frac{cos3x + 2cos3x \: cosx}{sin3x + 2sin3x \: cosx} \\ \\ = \displaystyle \sf \: \frac{cos3x(1 + 2cosx)}{sin3x(1 + 2cosx)} \\ \\ = \displaystyle \sf \: \frac{cos3x}{sin3x} \\ \\ = \displaystyle \sf \: cot3x$