Question

If $\cos \theta =\frac{1}{2}$ then the value of $\frac{2\sec \theta }{1+{\tan }^2\theta }$ is

• A. 1
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. 2

Answer 1

Answer: (A)

1

$cos\theta =\frac{1}{2}$

$\because cos{60}^{\circ }=\frac{1}{2}$

$\therefore \theta ={60}^{\circ }$

then $sec{60}^{\circ }=2$

$tan{60}^{\circ }=\sqrt{3}$

$\therefore \frac{2sec{60}^{\circ }}{1+ta{n}^2{60}^{\circ }}=\frac{2\times 2}{1+(\sqrt{3}{)}^2}$

$\Rightarrow \frac{4}{1+3}=\frac{4}{4}=1$