If θ is an acute angle such that sec2 θ = 3, then the value of tan2 θ−cosec2 θtan2 θ+cosec2 θis

Solution:

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 $(d)$