Question

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

Answer 1

Make use of the fact that as long as
`\sin(\theta) \\e 0` and
`\cos(\theta) \\e 0`:

`\displaystyle \tan(\theta) = (\sin(\theta))/(\cos(\theta))`.

`\displaystyle \csc(\theta) = (1)/(\sin(\theta))`.

`\sin^(2)(\theta) + \cos^(2)(\theta) = 1`.

Explanation:

Assume that
`\sin(\theta) \\e 0` and
`\cos(\theta) \\e 0`.

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

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

Since
`\cos(\theta) \\e 0`:

`\displaystyle 1 = (\cos^(2)(\theta))/(\cos^(2)(\theta))`.

Substitute this equality into the expression:

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

By the Pythagorean identity,
`\sin^(2)(\theta) + \cos^(2)(\theta) = 1`. Rearrange this identity to obtain:

`\sin^(2)(\theta) = 1 - \cos^(2)(\theta)`.

Substitute this equality into the expression:

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

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

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