Question

If $A+B={90}^o$ and $\sec A=\frac{5}{3}$, then find the value of $\csc B$.

Answer 1

$secA=\frac{5}{3}$

So $cosA=\frac{3}{5}$

$sin(90-A)=\frac{3}{5}$

$sinB=\frac{3}{5}[\because A+B={90}^{\circ }]$

so $cosecB=\frac{5}{3}$