Question

If $cosec\Theta =\sqrt{5}$, then

find the value of $2+\frac{1}{{\sin }^2\Theta }-\frac{{\cos }^2\Theta }{{\sin }^2\Theta }$

Answer 1

$cosec\Theta =\sqrt{5}$
$cosec\Theta =\frac{H}{P}=\sqrt{5}$
Thus, $H=\sqrt{5},P=1$
Using Pythagoras Theorem,
$H^2=P^2+B^2$
$(\sqrt{5}{)}^2=1^2+B^2$
$5=1+B^2$
$B=2$
Now, $2+\frac{1}{{\sin }^2\Theta }-\frac{{\cos }^2\Theta }{{\sin }^2\Theta }$
= $2+{\csc }^2\Theta -{\cot }^2\Theta$
= $2+(\frac{H}{P}{)}^2-(\frac{B}{P}{)}^2$
= $2+(\frac{\sqrt{5}}{1}{)}^2-(\frac{2}{1}{)}^2$
= $2+5-4$
= $3$