If x is an acute angle and tan x

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 $\operatorname{cosec}^{2} x$ :

$\frac{1-\tan ^{2} x}{1+\tan ^{2} x}$

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

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

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