Given that tan θ =
Question:

Given that $\tan \theta=\frac{1}{\sqrt{3}}$, the value of $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ is

(a) − 1

(b) 1

(c) $\frac{1}{2}$

(d) $-\frac{1}{2}$

Solution:

Given: $\tan \theta=\frac{1}{\sqrt{3}}$

We have to find the value of the following expression

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$

Now $\tan \theta=\frac{1}{\sqrt{3}}$, so

perpendicular =1

base $=\sqrt{3}$

and

hypotenus $=\sqrt{(\text { base })^{2}+(\text { perpendicular })^{2}}$

$=\sqrt{(\sqrt{3})^{2}+1^{2}}$

$=\sqrt{3+1}$

$=2$

Therefore

$\operatorname{cosec} \theta=\frac{\text { hypotenus }}{\text { perpendicular }}$

$=\frac{2}{1}=2$

and

$\sec \theta=\frac{\text { hypotenus }}{\text { base }}$

$=\frac{2}{\sqrt{3}}$

Put the above two values in the given expression, we have

$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{(2)^{2}-\left(\frac{2}{\sqrt{3}}\right)^{2}}{(2)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}$

$=\frac{4-\frac{4}{3}}{4+\frac{4}{3}}$

$=\frac{1}{2}$

Hence the correct option is $(c)$

Administrator

Leave a comment

Please enter comment.
Please enter your name.