Sine in terms of Tangent

Theorem
Let $\theta$ be an angle such that $\cos \theta \ne 0$.

Then:

where $\sin$ denotes the sine function and $\tan$ denotes the tangent function.

Proof
We also have that:

In quadrant I, $\sin \theta$ and $\tan \theta$ are positive, so $\sin \theta = + \dfrac {\tan \theta} {\sqrt{1 + \tan ^2 \theta} }$.

In quadrant II, $\sin \theta$ is positive but $\tan \theta$ is negative, so $\sin \theta = - \dfrac {\tan \theta} {\sqrt{1 + \tan ^2 \theta} }$.

In quadrant III, $\sin \theta$ is negative but $\tan \theta$ is positive, so $\sin \theta = - \dfrac {\tan \theta} {\sqrt{1 + \tan ^2 \theta} }$.

In quadrant IV, $\sin \theta$ and $\tan \theta$ are negative, so $\sin \theta = + \dfrac {\tan \theta} {\sqrt{1 + \tan ^2 \theta} }$.

$\tan \theta$ is not defined when $\cos \theta = 0$.

Also see

 * Trigonometric Functions in terms of each other