If $ \cot \theta=\frac{3}{4} $, prove that $ \sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}} $

Given:

\( \cot \theta=\frac{3}{4} \)

To do:

We have to prove that \( \sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}} \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$, $\ cot\ \theta = cot\ A = \frac{3}{4}$.

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,

$cosec\ \theta=\frac{Hypotenuse}{Opposite}=\frac{AC}{BC}$

$ecs\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$cot\ \theta=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(3)^2+(4)^2$

$\Rightarrow AC^2=9+16$

$\Rightarrow AC=\sqrt{25}=5$

Therefore,

$cosec\ \theta=\frac{AC}{BC}=\frac{5}{4}$

$sec\ \theta=\frac{AC}{AB}=\frac{5}{3}$

Now,

Let us consider LHS,

$\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\sqrt{\frac{\frac{5}{3} -\frac{5}{4}}{\frac{5}{3} +\frac{5}{4}}}$

$=\sqrt{\frac{\frac{5( 4) -5( 3)}{12}}{\frac{5( 4) +5( 3)}{12}}}$

$=\sqrt{\frac{5( 4-3)}{5( 4+3)}}$

$=\sqrt{\frac{1}{7}}$

$=\frac{1}{\sqrt7}$

$=$ RHS

Hence proved. 

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Updated on: 10-Oct-2022

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