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.

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Updated on: 10-Oct-2022

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