Question

$sin{60}^0=\frac{tan{30}^0}{1+ta{n}^2{30}^0}$.

• A. True
• B. False

Answer 1

The correct option is B False

As we know that

$\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$

Put $A={30}^{\circ }$

$⟹\sin 2\left({30}^{\circ }\right)=\frac{2\tan {30}^{\circ }}{1+{\tan }^2{30}^{\circ }}$

$⟹\sin {60}^{\circ }=\frac{2\tan {30}^{\circ }}{1+{\tan }^2{30}^{\circ }}$

So the Given relation is $\text{False}$