# How do you determine csctheta given sintheta=2/5,0^circ<theta<90^circ?

Dec 16, 2016

$\csc \theta = \frac{5}{2}$

#### Explanation:

As $\sin \theta$ is positive and is $\frac{2}{5}$

$\csc \theta$ being inverse of $\sin \theta$ i.e. $\csc \theta = \frac{1}{\sin} \theta$

= $\frac{1}{\frac{2}{5}} = 1 \times \frac{5}{2} = \frac{5}{2}$

Note that all the trigonometric ratios in range ${0}^{o} < \theta < {90}^{o}$ are positive, but this does not matter as if $\sin \theta$ is positive, $\csc \theta = \frac{1}{\sin} \theta$ too will have same sign i.e. it will be positive.