Question

plzz solve this question....​

Answer 1

Answer:

Step-by-step explanation:

Alright, lets get started.

The given equation is :

Starting from the left side

$tanx.tan(\frac{\pi }{3}-x).tan(\frac{\pi }{3}+x)$

$tanx.\frac{tan\frac{\pi }{3}-tanx }{1+tan\frac{\pi }{3}tanx}.\frac{tan\frac{\pi }{3}+tanx }{1-tan\frac{\pi }{3}tanx}$

Plugging the value of $tan(\frac{\pi }{3})$ as $\sqrt{3}$

$tanx.\frac{\sqrt{3} -tanx }{1+\sqrt{3} tanx}.\frac{\sqrt{3} +tanx }{1-\sqrt{3}tanx}$

$tanx . \frac{3-tan^2x}{1-3tan^2x}$

$\frac{3tanx-tan^3x}{1-3tan^2x}$  ( this is the formula of tan3x)

$tan(3x)$

Right side

Hence proved ...... Answer

Hope it will help :)