Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Asked by Pragya Singh | 1 year ago |  87

Solution :-

We know that,

sinA+sinB=2sin$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

LHS = cot 4x (sin 5x + sin 3x)

$$\dfrac{cot4x}{sin4x}[2sin(\dfrac{5x+3x}{2})cos(\dfrac{5x+3x}{2})]$$

$$( \dfrac{cos4x}{sin4x})[2sin4xcosx]$$

=2cos4xcosx

Now also ,we know that,

sinA-sinB=2cos$$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

R.H.S.= cotx(sin5x-sin3x)

$$\dfrac{cosx}{sinx}[2cos(\dfrac{5x+3x}{2})sin(\dfrac{5x-3x}{2})]$$

$$( \dfrac{cosx}{sinx})[2cos4xsinx]$$

=2cos4xcosx

Therefore , we can conclude that,

L.H.S.=R.H.S.

Hence proved.

Answered by Abhisek | 1 year ago

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