Question

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

Answer 1

Answer:

Answer for the given problem is given

Answer 2

Given:-

• $\large{\sf{\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}}}$

To Prove:-

• $\large{\sf{\frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}=\cot3x}}$

Solution:-

$\boxed{\frak{Using\: Identities}}\blue\bigstar$

$\sf{cos\: x + cos\: y = 2cos\: x+y/2 cos\: x-y/2}$

$\sf{sin \:x + sin \:y = 2sin\: x+y/2 cos\: x-y/2}$

$:\implies\sf{\dfrac{2cos\dfrac{4x+2x}{2}\:cos\dfrac{4x-2x}{2}+cos\:3x}{2sin\dfrac{4x+2x}{2}\:cos\dfrac{4x-2x}{2}}}$

$:\implies\sf{\dfrac{2cos\dfrac{{\not{6}}x}{{\not{2}}}\:cos\dfrac{{\not{2}}x}{{\not{2}}}+cos\:3x}{2sin\dfrac{{\not{6x}}}{{\not{2}}}\:cos\dfrac{{\not{2}}x}{{\not{2}}}}}$

$:\implies\sf{\dfrac{2\:cos\:3x\:cosx+cos\:3x}{2\:sin\:3x\:cosx+sin\:3x}}$

By Taking Common:

$:\implies\sf{\dfrac{cos\:3x{\cancel{(2\:cosx+1)}}}{sin\:3x{\cancel{(2\:cosx+1)}}}}$

$:\implies\sf{\dfrac{cos\:3x}{sin\:3x}}$

$:\implies\underline{\boxed{\frak{cot\:3x}}}\pink\bigstar$

• Hence, proved.

Thank you!

@itzshivani