Please help in finding the values

Mathematics

1 answer:

PIT_PIT [208]3 years ago

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Answer:umm

Step-by-step explanation:

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

Oxana [17]

Answer:

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

$\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

$\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}$ .

$\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ .

Step-by-step explanation:

Assume that $\sin(\theta) \ne 0$ and $\cos(\theta) \ne 0$ .

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 $\cos(\theta)$ .

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

Since $\cos(\theta) \ne 0$ :

$\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}$ .

Substitute this equality into the expression:

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

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:

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

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

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