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