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If \( \tan \theta=\frac{1}{\sqrt{7}} \), show that \( \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4} \)
Given:
$tan\ \theta = \frac{1}{\sqrt7}$.
To do:
We have to show that \( \frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{3}{4} \).
Solution:
Let, in a triangle $ABC$ right-angled at $B$, $\ tan\ \theta = tan\ A = \frac{1}{\sqrt7}$.
We know that,
In a right-angled triangle $ABC$ with right angle at $B$,
By Pythagoras theorem,
$AC^2=AB^2+BC^2$
By trigonometric ratios definitions,
$cosec\ \theta=\frac{Hypotenuse}{Opposite}=\frac{AC}{BC}$
$ecs\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$
$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$
Here,
$AC^2=AB^2+BC^2$
$\Rightarrow AC^2=(\sqrt7)^2+(1)^2$
$\Rightarrow AC^2=7+1$
$\Rightarrow AC=\sqrt{8}=2\sqrt2$
Therefore,
$cosec\ \theta=\frac{AC}{BC}=\frac{2\sqrt2}{1}=2\sqrt2$
$sec\ \theta=\frac{AC}{AB}=\frac{2\sqrt2}{\sqrt7}$
Now,
Let us consider LHS,
$\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}=\frac{\left( 2\sqrt{2}\right)^{2} -\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^{2}}{\left( 2\sqrt{2}\right)^{2} +\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^{2}}$
$=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$
$=\frac{48}{64}$
$=\frac{3}{4}$
$=$ RHS
Hence proved.