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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.