Prove that  $$\dfrac{sin5x+sin3x}{cos5x+cos3x}=tan4x$$

Asked by Pragya Singh | 1 year ago |  67

##### Solution :-

We know that

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

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

$$\dfrac{sin5x+sin3x}{cos5x+cos3x}=tan4x$$

Now ,

$$L.H.S.= \dfrac{sin5x+sin3x}{cos5x+cos3x}$$

$$\dfrac{2sin(\dfrac{5x+3x}{2})cos(\dfrac{5x-3x}{2})}{2cos(\dfrac{5x+3x}{2}).cos(\dfrac{5x-3x}{2})}$$

$$\dfrac{2sin4xcosx}{2cos4xcosx}$$

Further computing we have,

L.H.S=tan4x = R.H.S.

Answered by Abhisek | 1 year ago

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