Question

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

Answer 1

Given, cot θ = 3/4
We know that

$\cot \theta$= $\frac{adjacentside}{oppositeside}=\frac{3}{4}$
$Hypotenus{e}^2=adjacentsid{e}^2+oppositesid{e}^2$
=>$3^2+4^2$
=>$Hypotenus{e}^2=5^2$
=>$Hypotenuse=5$

$sec\theta =\frac{hypotenuse}{adjacentside}=\frac{5}{3}$
$cosec\theta =\frac{hypotenuse}{oppositeside}=\frac{5}{4}$

Now we put the values in the equation:

$\sqrt{\frac{sec\theta -cosec\theta }{sec\theta +cosec\theta }}$=$\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}}$
=>$\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}$
=>$\sqrt{\frac{5}{35}}$
=>$\frac{1}{\sqrt{7}}$
Hence proved