Question

If $\theta ={30}^0$ then , match the below expressions with there correct value .

• A. 3
• B. 2
• C. 32

Answer 1

Since, $\theta ={30}^0$
Case(1):
So,
Since $\tan {30}^0=\frac{1}{\sqrt{3}}$
$\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }$
= $\frac{1+{\tan }^2{30}^0}{1-{\tan }^2{30}^0}$
= $\frac{1+\frac{1}{3}}{1-\frac{1}{3}}$
= $\frac{\frac{4}{3}}{\frac{2}{3}}$
= 2

Case(2):
$\frac{2\tan \theta }{1-{\tan }^2\theta }$
= $\frac{2\tan {30}^0}{1-{\tan }^2{30}^0}$
=$\frac{2\times \frac{1}{\sqrt{3}}}{1-\frac{1}{3}}$
= $\sqrt{3}$

Case(3):
$\frac{2\tan \theta }{1+{\tan }^2\theta }$
= $\frac{2\tan {30}^0}{1+{\tan }^2{30}^0}$
=$\frac{2\times \frac{1}{\sqrt{3}}}{1+\frac{1}{3}}$
= $\frac{\sqrt{3}}{2}$