Proving $\tan\left(x+\frac{\pi}{3}\right) = \frac{4\tan x+\sqrt{3}\sec^2x}{\sec^2x-4\tan^2x}$

Question

I know this is a kind of elementary question, but I'm stuck with this one.

$$\tan\left(x + \frac{\pi}{3}\right) = \frac{4\tan x + \sqrt{3} \sec^2x}{\sec^2x-4\tan^2x}$$

Expanding the left side, we have $$\frac{\tan x + \sqrt{3}}{1-\sqrt{3}\tan x}$$ but after this, I'm stuck, I have already tried the usual stuff of multiplying by 1 etc.

Could someone give me a subtle hint?

Answer 1

Let me start from what you already got:

$\frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x}$

$ = \frac{(\tan x + \sqrt{3})(1 + \sqrt{3} \tan x)}{(1 - \sqrt{3} \tan x)(1 + \sqrt{3} \tan x)}$

$ = \frac{\tan x + \sqrt{3} \tan^2 x + \sqrt{3} + 3 \tan x}{1-3 \tan^2 x}$

$ = \frac{4 \tan x + \sqrt{3} \sec^2 x - \sqrt{3} + \sqrt{3}}{4-3 \sec^2 x}$

$ = \frac{4 \tan x + \sqrt{3} \sec^2 x}{4(\sec^2 x - \tan^2 x)-3 \sec^2 x}$

$ = \frac{4 \tan x + \sqrt{3} \sec^2 x}{\sec^2 x - 4\tan^2 x}$

Answer 2

Now, $$\frac{4\tan x + \sqrt{3} \sec^2x}{\sec^2x-4\tan^2x}=\frac{4\tan x + \sqrt{3} (1+\tan^2x)}{1+\tan^2x-4\tan^2x}=$$ $$=\frac{(\tan{x} + \sqrt{3})(1+\sqrt3\tan{x})}{(1-\sqrt{3}\tan{x})(1+\sqrt3\tan{x})}=\frac{\tan{x} + \sqrt{3}}{1-\sqrt{3}\tan{x}}=\frac{\tan{x} + \tan\frac{\pi}{3}}{1-\tan\frac{\pi}{3}\tan{x}}=\tan\left(x+\frac{\pi}{3}\right).$$