Question

Express $\tan \theta$ in terms of $\sin \theta$.

Answer 1

Using properties of trigonometric functions:

We know that, $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Replace the $\cos \theta$ in terms of $\sin \theta$ in $\tan \theta =\frac{\sin \theta }{\cos \theta }$, we get

$$\begin{gathered} \Rightarrow \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }} \end{gathered}$$

Hence, $\tan \theta$ in terms of $\sin \theta$ can be expressed as $\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$ .