We need to use the following formulas:

a) $\displaystyle \cos 2x = 2\cos^2x - 1 = 1 - 2\sin^2x = \cos^2x - \sin^2x$

b) $\displaystyle \sin^2x + \cos^2 x = 1$

c) $\displaystyle \csc x = \frac{1}{\sin x}$

We can simplify $\displaystyle f\left ( x\right )$ as follows:

$\displaystyle f\left ( x\right ) = \frac{\cos^2x - \sin^2 x}{1 - \cos^2 x} = \frac{1 - 2\sin^2x}{\sin^2x} = \frac{1}{\sin^2x} - 2 = \csc^2x - 2$

To solve this problem, we need to use the formula:

$\displaystyle \cos\left ( 2x\right ) = 1 - 2\sin^2\left (x\right )$

Substituting $\displaystyle \sin \left ( x\right ) = a$, we get

$\displaystyle \cos\left ( 2x\right ) = 1 - 2a^2$

In order to prove this trigonometric equation we can work with either the left or right side of the equation and attempt to make them equal. We will choose to work with the left side of the equation. First we separate the fractional term:

$\displaystyle \frac{2}{1-\cos 2A} \cos A = \csc A \cot A$

We separated the fractional term because we notice we have a double angle. Recalling our trigonometric identities, the fractional term is the inverse of the power reducing formula for sine.

$\displaystyle \frac{1}{\sin^{2}A}\cos A = \csc A \cot A$

Now separating out the sine terms:

$\displaystyle \frac{1}{\sin A}\frac{\cos A}{\sin A} = \csc A \cot A$

Now recalling the basic identities:

$\displaystyle \csc A \cot A = \csc A \cot A$

Using the trigonometric identities we have proven that the equation is true.

We choose which side to work with in the given equation. Selecting the right hand side since it contains a double angle we attempt to use the double angle formula to determine the equivalence:

$\displaystyle \csc A \sec A = \frac{2}{2\sin A \cos A}$

Next we reduce and split the fraction as follows:

$\displaystyle \csc A \sec A = \frac{1}{\sin A}\frac{1}{\cos A}$

Recalling the basic identities:

$\displaystyle \csc A \sec A = \csc A \sec A$

This proves the equivalence.

The formula for a doubled angle with sine is $\displaystyle \sin(2a)=2\sin (a)\cos(a)$

Plug in our given value and solve.

$\displaystyle \sin(2*\frac{\pi}{3})=2\sin (\frac{\pi}{3})\cos(\frac{\pi}{3})$

$\displaystyle \sin(2*\frac{\pi}{3})=2*\frac{\sqrt{3}}{2}*\frac{1}{2}$

Combine our terms.

$\displaystyle \sin(2*\frac{\pi}{3})=\frac{2\sqrt{3}}{4}$

$\displaystyle \sin(2*\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Write the Pythagorean Identity.

 $\displaystyle sin^2(\theta)+cos^2(\theta)=1$

Substitute the value of $\displaystyle sin(\theta)= \frac{2}{3}$ and solve for $\displaystyle (cos\theta)$.

$\displaystyle (\frac{2}{3})^{2}+cos^2(\theta)=1$

$\displaystyle cos^2(\theta)=1-\frac{4}{9}= \frac{5}{9}$

$\displaystyle cos(\theta)=\pm \sqrt \frac{5}{9}$

Since the $\displaystyle (cos\theta)$ must be less than zero, choose the negative sign.

$\displaystyle cos(\theta)=-\frac{\sqrt{5}}{3}$

Write the double-angle identity of $\displaystyle cos(2\theta)$.

$\displaystyle cos(2\theta)= cos^2(\theta)-sin^2(\theta)$

Substitute the known values.

$\displaystyle (-\frac{\sqrt{5}}{3})^2-(\frac{2}{3})^2= \frac{5}{9}-\frac{4}{9}=\frac{1}{9}$

We can exploit the following trigonometric identity:

$\displaystyle sec^2 X = tan^2X + 1 = (0.83)^2 + 1 = 1.6889$

Then we can do:

$\displaystyle cos^2X = \frac{1}{sec^2X} = \frac{1}{1.6889} = 0.5921$

With this value we can conveniently find our solution to be:

$\displaystyle cos \ 2X = 2cos^2 X -1 = 2 * 0.5921 - 1 = 0.1842 \approx 0.18$

The key here is to double-angle identity for $\displaystyle sin(2ax)$ to simplify the function.

$\displaystyle 2sin(ax)cos(ax) = sin(2ax)$

In this case, $\displaystyle a = \pi$, which means...

$\displaystyle 2sin(\pi x)cos(\pi x) = sin(2\pi x)$

From there, we can use the fact that the period of $\displaystyle sin(bx)$ or $\displaystyle cos(bx)$ is $\displaystyle \frac{2\pi}{b}$. Consequently,

$\displaystyle \frac{2\pi}{2\pi} = 1$

Since $\displaystyle \sin 2x = 2\sin x\cos x$

and $\displaystyle \sin (4x) = \sin (2(2x))$,

then $\displaystyle \sin 4x = 2 \sin 2x \cos 2x$.

Here we have to use the double-angle identities for both sine and cosine,

$\displaystyle \sin 2x = 2\sin x\cos x$ and $\displaystyle \cos 2x = \cos^2 x -\sin^2 x$.

Using these identities:

$\displaystyle 2 \sin 2x \cos 2x = 2(2\sin x \cos x)(\cos^2 x-\sin^2 x)$

Using the distributive property:

$\displaystyle 4\sin x \cos^3 x - 4\sin^3 x \cos x$

We need to use the following formulas:

a) $\displaystyle \sec x = \frac{1}{\cos x}$

b) $\displaystyle \sin^2x = \frac{1}{2}\left ( 1-\cos 2x\right )$

c) $\displaystyle \cos^2x = \frac{1}{2}\left ( 1+\cos 2x\right )$

d) $\displaystyle \tan x = \frac{\sin x}{\cos x}$

e) $\displaystyle \tan^2x + 1 = \sec^2x$

We can simplify $\displaystyle f\left ( x\right )$ as follows:

$\displaystyle f\left ( x\right ) = \frac{1}{\cos^2x} - \frac{1-\cos2x}{1 + \cos2x} = \sec^2x - \left ( \frac{2sin^2x}{2cos^2x} \right ) = \sec^2x - \tan^2x = 1$