Since cotangent is positive and sine is negative, alpha must be in quadrant III.  $\displaystyle \cot \alpha = \frac{x}{y}$ then implies that $\displaystyle (x,y)=(-5,-1)$ is a point on the terminal side of alpha. 

$\displaystyle r=\sqrt{(x^2+y^2)}=\sqrt{26}$

$\displaystyle \sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}$

Since cosecant is negative, theta must be in quadrant III or IV. 

Since tangent is positive, it must be in quadrant I or III. 

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between $\displaystyle \pi$ and $\displaystyle \frac{3\pi}{2}$.

Secant is defined to be the ratio of $\displaystyle r$ to $\displaystyle x$ where $\displaystyle r$ is the distance from the origin. 

The Pythagoreanr Triple 5, 12, 13 helps us realize that $\displaystyle r = 13$. 

Since $\displaystyle x = 12$, the answer is $\displaystyle \frac{13}{12}$.

Use the following trigonometric identity to solve this problem.

$\displaystyle \csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}$

Using the Pythagorean triple 3,4,5, it is easy to find $\displaystyle \cos x = \frac {4}{5}$.

Using the Pythagorean triple 5,12,13, it is easy to find $\displaystyle \sin y = \frac {12}{13}$.

So substituting all four values into the top equation, we get

$\displaystyle \csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}$

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$\displaystyle sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

Since $\displaystyle \csc\theta=\frac{1}{\sin\theta}$:

 $\displaystyle \frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

Secant is the reciprocal of cosine.

$\displaystyle \sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\displaystyle \sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\displaystyle \sec(\theta) = \frac{17}{15} = 1.13$

Cotangent is the reciprocal of tangent.

$\displaystyle \cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\displaystyle \cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\displaystyle \cot(\theta) = \frac{24}{18} = 1.33$

Rewrite $\displaystyle cot(\frac{\pi}{4})$ in terms of sine and cosine.

$\displaystyle cot(\frac{\pi}{4})= \frac{cos(\frac{\pi}{4})}{sin(\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$

Evaluate each term separately.

$\displaystyle sec(60)= \frac{1}{cos(60)}=\frac{1}{0.5}= 2$

$\displaystyle csc(45)= \frac{1}{sin(45)}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}=\sqrt2$

$\displaystyle sec(60)-csc(45)=2-\frac{2\sqrt2}{2}=2-\sqrt2$