Question

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

Answer 1

$cot\theta =\frac{3}{4}$, we need to show that $\sqrt{\frac{sec\theta -cosec\theta }{sec\theta +cosec\theta }}=\frac{1}{\sqrt{7}}$

$cot\theta =\frac{Adjacentside}{Oppositeside}$

Let x be the hypotenuse by applying Pythagoras theorem.

$A{C}^2=A{B}^2+B{C}^2{x}^2=4^2+3^2{x}^2=16+9x^2=25x=\sqrt{25}=5sec\theta =\frac{AC}{BC}=\frac{5}{3}cosec\theta =\frac{AC}{AB}=\frac{5}{4}$

On substitution we get,

$\sqrt{\frac{sec\theta -cosec\theta }{sec\theta +cosec\theta }}=\sqrt{\frac{\frac{5}{3}-\frac{5}{4}}{\frac{5}{3}+\frac{5}{4}}}=\sqrt{\frac{\frac{20-15}{12}}{\frac{20+15}{12}}}=\sqrt{\frac{\frac{5}{12}}{\frac{35}{12}}}=\sqrt{\frac{1}{7}}=\frac{1}{\sqrt{7}}$

Hence proved