If $ \sin \theta=\frac{3}{4} $, prove that $ \sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3} $

Given:

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

To do:

We have to prove that \( \sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\frac{\sqrt{7}}{3} \).

Solution:  

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

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,

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

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

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

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

Here,

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

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

$\Rightarrow AB^2=16-9$

$\Rightarrow AB=\sqrt{7}$

Therefore,

$cosec\ \theta=\frac{AC}{BC}=\frac{4}{3}$

$sec\ \theta=\frac{AC}{AB}=\frac{4}{\sqrt7}$

$cot\ \theta=\frac{AB}{BC}=\frac{\sqrt7}{3}$

Now,

Let us consider LHS,

$\sqrt{\frac{\operatorname{cosec}^{2} \theta-\cot ^{2} \theta}{\sec ^{2} \theta-1}}=\sqrt{\frac{\left(\frac{4}{3}\right)^{2} -\left(\frac{\sqrt{7}}{3}\right)^{2}}{\left(\frac{4}{\sqrt{7}}\right)^{2} -1}}$

$=\sqrt{\frac{\frac{16}{9} -\frac{7}{9}}{\frac{16}{7} -1}}$

$=\sqrt{\frac{\frac{16-7}{9}}{\frac{16-7}{7}}}$

$=\sqrt{\frac{\frac{9}{9}}{\frac{9}{7}}}$

$=\sqrt{1\times \frac{7}{9}}$

$=\sqrt{\frac{7}{9}}$

$=\frac{\sqrt{7}}{3}$

$=$ RHS

Hence proved.  

Tutorialspoint

e

Updated on: 10-Oct-2022

66 Views

Kickstart Your Career

Get certified by completing the course

Get Started