Question:
If $x$ is an acute angle and $\tan x=\frac{1}{\sqrt{7}}$, then the value of $\frac{\operatorname{cosec}^{2} x-\sec ^{2} x}{\operatorname{cosec}^{2} x+\sec ^{2} x}$ is
(a) 3/4
(b) 1/2
(c) 2
(d) 5/4
Solution:
(a) 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 $\frac{\operatorname{cosec}^{2} x-\sec ^{2} x}{\operatorname{cosec}^{2} x+\sec ^{2} x}$ by $\operatorname{cosec}^{2} x$ :
$\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$
$=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}$
$=\frac{6}{8}=\frac{3}{4}$