Prove that $$tan4x=\dfrac{4tanx(1-tan^2x)}{1-6tan^2x+tan^4x}$$

Asked by Abhisek | 1 year ago |  62

Solution :-

We know that

$$tan2A=\dfrac{2tanA}{1-tan^2A}$$

L.H.S. = tan4x

=tan2(2x)

$$\dfrac{2tan2x}{1-tan^2(2x)}$$

$$\dfrac{\dfrac{4tanx}{1-tan^2x}}{[1-\dfrac{4tan^2x}{(1-tan^2x})^2]}$$

$$tan4x=\dfrac{4tanx(1-tan^2x)}{1-6tan^2x+tan^4x}$$

Further computing, we obtain

$$\dfrac{\dfrac{4tanx}{1-tan^2x}}{[\dfrac{(1-tan^2x)^24tan^2x)}{(1-tan^2x)^2}]}$$

$$\dfrac{4tanx(1-tan^2x)}{1+tan^4x-2tan^2x-4tan^2x}$$

$$\dfrac{4tanx(1-tan^2x)}{1-6tan^2x+tan^4x}$$

= R.H.S.

Hence proved.

Answered by Pragya Singh | 1 year ago

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