If tan θ=17√, then cosec2 θ−sec2 θcosec2 θ+sec2 θ=

Solution:

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 $(d)$