Question

If $\sec \theta =2,$ evaluating $\frac{1-\tan \theta }{1+\tan \theta }$ gives

• A. $-\sqrt{3}$
• B. $\sqrt{3}+1$
• C. $2-\sqrt{3}$
• D. $\frac{\sqrt{3}+1}{2}$

Answer 1

Answer: (C)

$2-\sqrt{3}$

we know that, ${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow {\tan }^2\theta ={\sec }^2\theta -1$
$\Rightarrow {\tan }^2\theta =4-1$ $[given\sec \theta =2]$
$\Rightarrow \tan \theta =\sqrt{3}$
Now,$\frac{1-\tan \theta }{1+\tan \theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}\times \frac{1-\sqrt{3}}{1-\sqrt{3}}$
$=\frac{1+3-2\sqrt{3}}{1-3}$
$=\frac{4-2\sqrt{3}}{-2}$
$=\sqrt{3}-2$