If $\theta$ is an acute angle such that $\sec ^{2} \theta=3$, then the value of $\frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}$ is
(a) $\frac{4}{7}$
(b) $\frac{3}{7}$
(c) $\frac{2}{7}$
(d) $\frac{1}{7}$
Given that:
$\sec ^{2} \theta=3$
$\sec \theta=\sqrt{3}$
We need to find the value of the expression
$\frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}$
Since $\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$.So
$\Rightarrow$ Hypotenuse $=\sqrt{3}$
$\Rightarrow$ Base $=1$
$\Rightarrow$ Perpendicular $=\sqrt{3-1}$
$\Rightarrow$ Perpendicular $=\sqrt{2}$
Here we have to find: $\frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}$
$\Rightarrow \frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}=\frac{\frac{2}{1}-\frac{3}{2}}{\frac{2}{1}+\frac{3}{2}}$
$\Rightarrow \frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}=\frac{\frac{1}{2}}{\frac{7}{2}}$
$\Rightarrow \frac{\tan ^{2} \theta-\operatorname{cosec}^{2} \theta}{\tan ^{2} \theta+\operatorname{cosec}^{2} \theta}=\frac{1}{7}$
Hence the correct option is $(d)$