Question

If $\tan \left(\theta \right)=\frac{1}{\sqrt{7}}$, then $\frac{cose{c}^2\left(\theta \right)-{\sec }^2\left(\theta \right)}{cose{c}^2\left(\theta \right)+{\sec }^2\left(\theta \right)}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. $2$

Answer 1

The correct option is B

$\frac{3}{4}$

Explanation for the correct option

Given: $\frac{cose{c}^2\left(\theta \right)-{\sec }^2\left(\theta \right)}{cose{c}^2\left(\theta \right)+{\sec }^2\left(\theta \right)}$

$=\frac{{\displaystyle \frac{1}{{\sin }^2\left(\theta \right)}}-{\displaystyle \frac{1}{{\cos }^2\left(\theta \right)}}}{{\displaystyle \frac{1}{{\sin }^2\left(\theta \right)}}+{\displaystyle \frac{1}{{\cos }^2\left(\theta \right)}}}$

multiplying ${\sin }^2\left(\theta \right)$ to both numerator and denominator

$$\begin{gathered} =\frac{{\displaystyle \frac{{\sin }^2\left(\theta \right)}{{\sin }^2\left(\theta \right)}}-{\displaystyle \frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}}}{{\displaystyle \frac{{\sin }^2\left(\theta \right)}{{\sin }^2\left(\theta \right)}}+{\displaystyle \frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)}}} \\ =\frac{{\displaystyle 1-{\tan }^2\left(\theta \right)}}{{\displaystyle 1+{\tan }^2\left(\theta \right)}} \\ =\frac{{\displaystyle 1-{\left(\frac{1}{\sqrt{7}}\right)}^2}}{{\displaystyle 1+{\left(\frac{1}{\sqrt{7}}\right)}^2}}\left[\because \tan \left(\theta \right)=\frac{1}{\sqrt{7}}\right] \\ =\frac{{\displaystyle \frac{7-1}{7}}}{{\displaystyle \frac{7+1}{7}}} \\ =\frac{{\displaystyle 6}}{{\displaystyle 8}} \\ =\frac{3}{4} \end{gathered}$$

Hence, option B is correct.