Question

prove that : tan 4x = 4 tan x (1 – tan- x) ÷ 1-6 tan 2 x + tan 4 х​

Answer 1

Step-by-step explanation:

Given:

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

To Find:

To prove that $tan4x=\frac{4 tanx (1-tan^{2} x)}{1-6tan^{2} x+tan^4x}$              

Solution:

Given that,

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

Where,

Left Hand side $LHS=tan4x$

Right Hand side $RHS=\frac{4 tanx (1-tan^{2} x)}{1-6tan^{2} x+tan^4x}$

Now take LHS,

$tan4x=tan2(2x)$

         $=\frac{2tan2x}{1-tan^22x}$

         $=\frac{2(\frac{2tanx}{1-tan^2x}) }{1-(\frac{2tanx}{1-tan^2x})^2 }$

         $=\frac{\frac{4tanx}{1-tan^2x} }{1-\frac{4tan^2x}{(1-tan^2x)^2} }$

         $=\frac{\frac{4tanx}{1-tan^2x} }{\frac{(1-tan^2x)^2-4tan^2x}{(1-tan^2x)^2} }$

        $=\frac{4tanx(1-tan^2x)}{(1-tan^2x)^2-4tan^2x}$

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

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

Hence proved.

Answer:

Hence the expression $tan4x=\frac{4 tanx (1-tan^{2} x)}{1-6tan^{2} x+tan^4x}$ is proved.