Prove that $$\dfrac{sinx+sin3x}{cosx+cos3x}=tan2x$$

Asked by Abhisek | 1 year ago |  64

Solution :-

We know that

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

cosA+cosB=2cos$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

Now , L.H.S. $$\dfrac{sinx+sin3x}{cosx+cos3x}$$

$$\dfrac{2sin(\dfrac{x+3x}{2})cos(\dfrac{x-3x}{2})}{2cos(\dfrac{x+3x}{2})cos(\dfrac{x-3x}{2})}$$

$$\dfrac{sin2x}{cos2x}=tan2x$$

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

Hence proved

Answered by Pragya Singh | 1 year ago

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