Prove that:$$\dfrac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}=tan6x$$

Asked by Pragya Singh | 1 year ago |  116

##### Solution :-

We known that,

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

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

And L.H.S.=

$$\dfrac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}$$

Using the formula and simplifying,

$$\dfrac{[2sin6x.cosx]+[2sin6x.cos3x]}{[2cos6x.cosx]+[2cos6x.cos6x]}$$

$$\dfrac{2sin6x[cosx+cos3x]}{2cos6x[cosx+cos3x]}$$

=tan6x

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

Hence proved.

Answered by Abhisek | 1 year ago

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