Question

Given : $tanA=\frac{4}{3}$,
: $\frac{cosecA}{cotA-secA}$ is $-\frac{15}{m}$, m is

Answer 1

$\tan A=\frac{4}{3}$
$\tan A=\frac{P}{B}=\frac{4}{3}$
Using Pythagoras Theorem,
$H^2=P^2+B^2$
$H^2=4^2+3^2$
$H=5$
Now, $\frac{cosecA}{cotA-secA}$
= $\frac{\frac{H}{P}}{\frac{B}{P}-\frac{H}{B}}$
= $\frac{\frac{5}{4}}{\frac{3}{4}-\frac{5}{3}}$
= $\frac{\frac{5}{4}}{\frac{9-20}{12}}$
= $\frac{\frac{5}{4}}{\frac{-11}{12}}$
= $-\frac{15}{11}$