Question

If $\theta$ is an acute angle and $tan\theta =\frac{1}{\sqrt{7}}$, then the value of $\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }$ is

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

Answer 1

The correct option is A

$\frac{3}{4}$

$\frac{3}{4}$

We have:

$tan\theta =\frac{1}{\sqrt{7}}$

$\therefore ta{n}^2\theta =\frac{1}{7}$

Now, dividing the numerator and the denominator of $\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }$ by $cose{c}^2\theta :$

$\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$

$=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}$

$=\frac{6}{8}=\frac{3}{4}$