Answer:
This identity holds as long as [tex]\displaystyle \theta \ne k\, \pi + \frac{\pi}{2}[/tex] for all integer [tex]k[/tex].
For the proof, make use of the fact that:
[tex]\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}[/tex] (definition of tangents,) and
[tex]\cos(\theta) = \sqrt{1 - \sin^{2}(\theta)}[/tex] (Pythagorean identity,) which is equivalent to [tex]1 - \cos^{2}(\theta) = \sin^{2}(\theta)[/tex].
Step-by-step explanation:
Assume that [tex]\displaystyle \theta \ne k\, \pi + \frac{\pi}{2}[/tex] for all integer [tex]k[/tex]. This requirement ensures that the [tex]\tan(\theta)[/tex] on the left-hand side takes a finite value. Doing so also ensures that the denominator [tex]\sqrt{1 - \sin^2(\theta)}[/tex] on the right-hand side is non-zero.
Make use of the fact that [tex]\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}[/tex] to rewrite the left-hand side:
[tex]\begin{aligned} & \tan(\theta) \cdot \sin(\theta) \\ =&\; \frac{\sin({\theta})}{\cos({\theta})} \cdot \sin(\theta) \\ =&\; \frac{\sin^{2}(\theta)}{\cos(\theta)}\end{aligned}[/tex].
Apply the Pythagorean identity [tex]\sin^{2}(\theta) = 1 - \cos^{2}(\theta)[/tex] and [tex]\cos(\theta) = \sqrt{1 - \sin^{2}(\theta)}[/tex] to rewrite this fraction:
[tex]\begin{aligned} & \frac{\sin^{2}(\theta)}{\cos(\theta)}\\ =\; &\frac{1 - \cos^{2}(\theta)}{\cos(\theta)}\\ =\; & \frac{1 - \cos^{2}(\theta)}{\sqrt{1 - \sin^{2}(\theta)}}\end{aligned}[/tex].
Hence, [tex]\displaystyle \tan(\theta) \cdot \sin(\theta) = \frac{1 - \cos^{2}(\theta)}{\sqrt{1 - \sin^{2}(\theta)}}[/tex].
