Question

If $\cos \theta =\frac{2\sqrt{6}}{7},$ then the value of $\sqrt{\frac{cose{c}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}$ is

• A. $\frac{2\sqrt{6}}{7}$
  
• B. $\frac{5}{7}$
  
• C. $\frac{2\sqrt{6}}{5}$
  
• D. $\frac{2\sqrt{6}}{3}$
  

Answer 1

The correct option is C $\frac{2\sqrt{6}}{5}$


Given: $\cos \theta =\frac{2\sqrt{6}}{7}$

In a right triangle,

$\cos \theta =\frac{\text{Base (B)}}{\text{Hypotenuse (H)}}=\frac{2\sqrt{6}}{7}$

Let $B=2\sqrt{6}kandH=7k$ (where k is any positive integer)

By Pythagoras theorem,

$H^2=P^2+B^2$

$\Rightarrow (7k{)}^2=P^2+(2\sqrt{6}k{)}^2$

$\Rightarrow P^2=49k^2-24k^2=25k^2$

$\Rightarrow P=5k$

$\therefore \cot \theta =\frac{B}{P}=\frac{2\sqrt{6}k}{5k}=\frac{2\sqrt{6}}{5}$ ...(i)

Now, consider $\sqrt{\frac{cose{c}^2\theta -{\cot }^2\theta }{{\sec }^2\theta -1}}=\sqrt{\frac{1}{{\tan }^2\theta }}(\because 1+{\cot }^2\theta =cose{c}^2\theta and1+{\tan }^2\theta ={\sec }^2\theta )$

$=\sqrt{{\cot }^2\theta }$

$=\cot \theta$

$=\frac{2\sqrt{6}}{5}$ [From (i)]

Hence, the correct answer is option (c).