Question

if tanθ+secθ=x, prove that sinθ=(X^2-1)(x^2+1)

Answer 1

Conditions

if $\tan \theta +\sec \theta = x$, prove that $\sin \theta = (X^{\wedge}2 - 1)(x^{\wedge}2 + 1)$

Solution

We must prove the following:

$$\sin \theta = ((\tan \theta + \sec \theta)^2 - 1)((\tan \theta + \sec \theta)^2 + 1)$$

But this is not true for all $\theta$. For example, let's take $\theta = \pi i / 4$. Then:

$$\begin{array}{l} ((\tan \theta + \sec \theta)^2 - 1)((\tan \theta + \sec \theta)^2 + 1) = ((1 + \sqrt{2})^2 - 1)((1 + \sqrt{2})^2 + 1) \\ = (1 + \sqrt{2})^4 - 1 \approx 32.97 \end{array}$$

And this value must be equal to $\sin \theta$. But sinus is a bounded function in interval [-1,1]. So the equation is wrong.

The graph of $(\tan \theta + \sec \theta)^2 + 1$ is below:

As we see, the graph isn't bounded between -1 and 1 not only in 1 point.