How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)

Make use of the fact that [tex]\tan(\theta) = (\sin(\theta)) / (\cos(\theta))[/tex] and [tex]\csc(\theta) = (1) / (\sin(\theta))[/tex] to rewrite the given expression as a combination of [tex]\sin(\theta)[/tex] and [tex]\cos(\theta)[/tex].

[tex]\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}[/tex].

Since [tex]\cos(\theta) \ne 0[/tex]:

[tex]\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}[/tex].

Substitute this equality into the expression:

[tex]\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}[/tex].

By the Pythagorean identity, [tex]\sin^{2}(\theta) + \cos^{2}(\theta) = 1[/tex]. Rearrange this identity to obtain:

[tex]\sin^{2}(\theta) = 1 - \cos^{2}(\theta)[/tex].

Substitute this equality into the expression:

[tex]\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}[/tex].

Again, make use of the fact that [tex]\tan(\theta) = (\sin(\theta)) / (\cos(\theta))[/tex] to obtain the desired result:

[tex]\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}[/tex].