Question

If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is

Options

• 3/4

• 1/2

• 2

• 5/4

Answer 1

3/4

We have: 

\[\tan x = \frac{1}{\sqrt{7}}\]

\[ \therefore \tan^2 x = \frac{1}{7}\]

Now, dividing the numerator and the denominator of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\text{ by }{cosec}^2 x:\]
\[\frac{1 - \tan^2 x}{1 + \tan^2 x}\]
\[ = \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}}\]
\[ = \frac{6}{8} = \frac{3}{4}\]