Question

Given: $\tan A=\frac{4}{3}$ ,
$\frac{cosecA}{\cot A-\sec A}$ 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, $\frac{cosecA}{\cot A-\sec A}$

= $\frac{cosecA}{\cot A-\sec A}$

= $\frac{\frac{H}{P}}{\frac{B}{P}-\frac{H}{B}}$

= $\frac{\frac{5}{4}}{\frac{3}{4}-\frac{5}{3}}$

= $\frac{\frac{5}{4}}{\frac{9-20}{12}}$

= $-\frac{15}{11}$