Recall that \(\displaystyle sec(x)=\frac {1}{cos(x)}\), and that \(\displaystyle cos(-x)=cos(x)\)

Therefore:

\(\displaystyle sec(x)\cdot cos(-x)=sec(x)\cdot cos(x)=1\)

Since that term is eliminated, we have left:

\(\displaystyle -csc(-x)cot(-x)\)

Recall that

\(\displaystyle csc(-x)=-csc(x)\) 

\(\displaystyle cot(-x)=-cot(x)\)

Therefore:

\(\displaystyle -csc(-x)cot(-x)=-(-csc(x)\cdot-cot(x))=-csc(x)cot(x)\)

We can use the following trigonometric identity to help us in the calculation:

\(\displaystyle \small \tan(X)=\frac{1-\cos2X}{\sin2X}\)

We plug in \(\displaystyle \small X=\frac{\pi}{12}\) to get

\(\displaystyle \tan\left(\frac{\pi}{12}\right)=\frac{1-\cos \pi/6}{\sin\pi/6}=\frac{1-\frac{\sqrt{3}}{2}}{1/2}\)

\(\displaystyle \small =2-\sqrt{3}\).

We can use the trigonometric identity, 

\(\displaystyle \small \tan(x)=\frac{1-\cos2x}{\sin2x}\)

along with the fact that

\(\displaystyle \small \cot(x)=\frac{1}{\tan(x)}\)

to compute \(\displaystyle \small \cot(\pi/8)\).

We have

\(\displaystyle \small \small \cot(\pi/8)=\frac{1}{\tan \pi/8}=\frac{\sin \pi/4}{1-\cos\pi/4}\)

\(\displaystyle \small =\frac{\sqrt{2}/2}{1-\sqrt{2}/2}=\frac{\sqrt{2}}{2-\sqrt{2}}\)

When trying to identify equivalent equations that use trigonometric functions it is important to recall the general formula and understand how the terms affect the translations.

The general formula for sine is as follows.

\(\displaystyle f(x)=a\sin(bx-c)+d\) where \(\displaystyle a\) is the amplitude, \(\displaystyle b\) is used to find the period of the function \(\displaystyle \frac{2\pi}{|b|}\), \(\displaystyle c\) represents the phase shift \(\displaystyle \frac{c}{b}\), and \(\displaystyle d\) is the vertical shift.

This is also true for,

\(\displaystyle f(x)=a\cos(bx-c)+d\).

Looking at the possible answer choices lets first focus on the ones containing sine.

\(\displaystyle f(x)=\sin (x)+\frac{\pi}{2}\) has a vertical shift of \(\displaystyle \frac{\pi}{2}\) therefore it is not an equivalent function as it is moving the original function up.

\(\displaystyle f(x)=\sin\left(x-\frac{\pi}{2}\right)\) has a phase shift of \(\displaystyle \frac{\pi}{2}\) therefore it is not an equivalent function as it is moving the original function to the right.

Now lets shift our focus to the answer choices that contain cosine.

\(\displaystyle f(x)=\cos(x)-\frac{\pi}{2}\) has a vertical shift down of \(\displaystyle \frac{\pi}{2}\) units. This will create a graph that has a range that is below the \(\displaystyle x\)-axis. It is important to remember that \(\displaystyle f(x)=sin(x)\) has a range of \(\displaystyle (-1,1)\). Therefore this cosine function is not an equivalent equation.

\(\displaystyle f(x) = \cos\left(x-\frac{\pi}{2}\right)\) has a phase shift to the right \(\displaystyle \frac{\pi}{2}\) units. Plugging in some values we see that,

\(\displaystyle f(0)=\cos\left(0-\frac{\pi}{2}\right)=0\), 

\(\displaystyle f\left(\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\pi}{2}\right)=\cos(0)=1\).

Now, looking back at our original function and plugging in those same values of \(\displaystyle x=0\) and \(\displaystyle x=1\) we get,

\(\displaystyle f(0)=sin(0)=0\),

\(\displaystyle f\left(\frac{\pi}{2}\right)=sin\left(\frac{\pi}{2}\right)=1\).

Since the function values are the same for each of the input values, we can conclude that \(\displaystyle f(x) = \cos\left(x-\frac{\pi}{2}\right)\) is equivalent to \(\displaystyle f(x)=sin(x)\).

First, factor \(\displaystyle sin^4A-cos^4A\) into their simplified form.

\(\displaystyle sin^4A-cos^4A=(sin^2A+cos^2A)(sin^2A-cos^2A)\)

The identity \(\displaystyle sin^2A+cos^2A\) equals to 1.

Factor \(\displaystyle sin^2A-cos^2A\).

\(\displaystyle sin^2A-cos^2A=(sinA+cosA)(sinA-cosA)\)

Since:

\(\displaystyle sin A + cos A=x\) 

\(\displaystyle sin A-cosA=y\)

Substitute the values of the simplified equation.

\(\displaystyle (sinA+cosA)(sinA-cosA)=xy\)

In order to find the exact value of \(\displaystyle \sin165^\circ\) we can use the half angle formula for sin, which is  

\(\displaystyle sin\left(\frac{\alpha}{2}\right) = \pm\sqrt\frac{1-cos\alpha}{2}\).

This way we can plug in a value for alpha for which we know the exact value. \(\displaystyle 165^\circ\) is equal to \(\displaystyle 330^\circ\) divided by two, and so we can plug in \(\displaystyle 330^\circ\) for the alpha above.

The cosine of \(\displaystyle 330^\circ\) is \(\displaystyle \frac{\sqrt3}{2}\).

Therefore our final answer becomes,

\(\displaystyle \pm\sqrt\frac{\left(1-\frac{\sqrt(3)}{2}\right)}{2}\).

\(\displaystyle \frac{\sin x}{\cos x}=\tan x\)

\(\displaystyle \frac{1}{\sin x}=\csc x\)

Given these identities...

\(\displaystyle 1-\tan x(\csc x)\)

\(\displaystyle 1- \frac{\sin x}{\cos x}(\frac{1}{\sin x})\)

\(\displaystyle 1-\frac{1}{\cos x}\)

\(\displaystyle 1- \sec x\)

First simplify the fraction 

\(\displaystyle \frac{\sin (x)}{1-\cos (x)}\) 

by multiplying it by its conjugate

\(\displaystyle \frac{1+\cos (x)}{1+\cos (x)}\).

After doing so, continue simplying:

\(\displaystyle \\ \frac{\sin(x)(1+\cos(x))}{(1-\cos(x))(1+\cos (x))}- \cot(x)\\ \\=\frac{\sin(x)(1+\cos(x))}{1-\cos^2(x)}-\cot(x)\\ \\=\frac{\sin(x)(1+\cos(x))}{\sin^2(x)}-\cot(x)\\ \\=\frac{1+\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\ \\=\frac{1}{\sin(x)}\\ \\=\csc (x)\) 

\(\displaystyle \sin^2 x + \cos^2 x = 1\)

\(\displaystyle 1+ \cot^2x= \csc^2 x\)

Given the above identities:

\(\displaystyle \frac{(\sin ^2 x+ \cos^2 x)}{(1+ \cot^2 x)}=\frac{1}{\csc^2 x}=\sin^2x\)

\(\displaystyle \sin^2x+\cos^2x=1\)

and

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

Therefore...

\(\displaystyle \frac{1-\cos ^2x}{\sin ^2x}+(\sin ^2x + \cos^2 x)\)

\(\displaystyle =\frac{\sin^2x}{\sin^2x}+(\sin^2x+\cos^2x)\)

\(\displaystyle =1+(\sin ^2x+\cos ^2x)\)

\(\displaystyle =1+1=2\)