Question

If $\cos A=\frac{9}{41};$ find the value of:$\frac{1}{{\sin }^{2}A}-{\cot }^{2}A$

Answer 1

$\cos A=\frac{9}{41}$
$\cos A=\frac{B}{H}=\frac{9}{41}$
Using Pythagoras Theorem,
$H^2=P^2+B^2$
${41}^2=P^2+9^2$
$P=40$
Now, $\frac{1}{{\sin }^{2}A}-{\cot }^{2}A$
= $cose{c}^{2}A-{\cot }^{2}A$
= $(\frac{H}{P}{)}^2-(\frac{B}{P}{)}^2$
= $(\frac{41}{40}{)}^2-(\frac{9}{40}{)}^2$
= $\frac{1600}{1600}$
= $1$