Question

Given: $\sec A=\frac{29}{21}$ ,
$\sin A-\frac{1}{\tan A}$ is $-\frac{209}{m}$, m is

Answer 1

$\sec A=\frac{29}{21}$

$\sec A=\frac{H}{B}=\frac{29}{21}$

Thus, $H=29,B=21$

Using Pythagoras Theorem,

$H^2=P^2+B^2$

${29}^2=P^2+{21}^2$

$P^2=841-441$

$P=20$

Now, $\sin A-\frac{1}{\tan A}=\frac{P}{H}-\frac{1}{\frac{P}{B}}$

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

= $\frac{20}{29}-\frac{21}{20}$

= $\frac{400-609}{580}$

= $-\frac{209}{580}$