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Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that cos4x+cos3x+cos2xsin4x+sin3x+sin2x=cot 3x\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x}=cot\,3x.

Answer

It is known that

L.H.S=cos4x+cos3x+cos2xsin4x+sin3x+sin2xL.H.S=\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x}

=(cos4x+cos2x)+cos3x(sin4x+sin2x)+sin3x=\frac{(cos4x+cos2x)+cos3x}{(sin4x+sin2x)+sin3x}

=2cos(4x+2x2)cos(4x−2x2)+cos3x2sin(4x+2x2)cos(2x−2x2)+sin3x=\frac{2cos(\frac{4x+2x}{2})cos(\frac{4x-2x}{2})+cos3x}{2sin(\frac{4x+2x}{2})cos(\frac{2x-2x}{2})+sin3x}

[cosA−cosB=2cos(A+B2)cos(A−B2),sinA−sinB=2sin(A+B2)cos(A−B2)][cosA-cosB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2}),sinA-sinB=2sin(\frac{A+B}{2})cos(\frac{A-B}{2})]

=2cos3xcosx+cos3x2sin3xcosx+sin3x=\frac{2cos3xcosx+cos3x}{2sin3xcosx+sin3x}

=cos3x(2cos+1)sin3x(2cosx+1)=\frac{cos3x(2cos+1)}{sin3x(2cosx+1)}

=cot 3x=R.H.S.\text{=cot\,3x=R.H.S.}