If $\tan \theta=\frac{1}{\sqrt{7}}$, then $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=$
(a) $\frac{5}{7}$
(b) $\frac{3}{7}$
(c) $\frac{1}{12}$
(d) $\frac{3}{4}$
Given that:
$\tan \theta=\frac{1}{\sqrt{7}}$
We are asked to find the value of the following expression
$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$
Since $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow$ Perpendicular $=1$
$\Rightarrow$ Base $=\sqrt{7}$
$\Rightarrow$ Hypotenuse $=\sqrt{1+7}$
$\Rightarrow$ Hypotenuse $=\sqrt{8}$
We know that $\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$ and $\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$
We find:
$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$
$=\frac{\left(\frac{\sqrt{8}}{1}\right)^{2}-\left(\frac{\sqrt{8}}{\sqrt{7}}\right)^{2}}{\left(\frac{\sqrt{8}}{1}\right)^{2}+\left(\frac{\sqrt{8}}{\sqrt{7}}\right)^{2}}$
$=\frac{\frac{8}{1}-\frac{8}{7}}{\frac{8}{1}+\frac{8}{7}}$
$=\frac{\frac{48}{7}}{\frac{64}{7}}$
$=\frac{3}{4}$
Hence the correct option is $(d)$