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.

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

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

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

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

Find the ratio of Cosine and take the reciprocal.

\(\displaystyle \cos\theta=\frac{AC}{CB}\rightarrow \sec\theta=\frac{1}{\frac{AC}{CB}}=\mathbf{\frac{CB}{AC}}\) 

Recall that \(\displaystyle sec(x) = \frac{1}{cos(x)}\) and \(\displaystyle csc(x) = \frac{1}{sin(x)}\).

Rewrite the expression.

\(\displaystyle sec(30)-csc(60) = \frac{1}{cos(30)}-\frac{1}{sin(60)}\)

The value of \(\displaystyle cos(30) = \frac{\sqrt3}{2}\) and \(\displaystyle sin(60) =\frac{ \sqrt{3}}{2}\).

Since these values are similar, our resulting answer is zero upon substitution.

\(\displaystyle \frac{1}{\frac{\sqrt3}{2}}-\frac{1}{\frac{\sqrt3}{2}} = 0\)

The answer is:  \(\displaystyle 0\)

Dropping the altitude creates two special right triangles as shown in the diagram.  Use the area formula of a triangle to get

\(\displaystyle A = \frac {1}{2}bh = \frac {1}{2} \cdot (6 \sqrt{3} + 6) \cdot 6 = 18\sqrt{3} + 18\)

First, realize that the angles given are from due north, which means you need to find the complements to find the interior angles of the triangle.  This triangle happens to be a right triangle, so the fast way to compute the distances is using trigonometry.

\(\displaystyle \sin 50^\circ = \frac{AF}{2}\)

\(\displaystyle AF \approx 1.53 miles\)

\(\displaystyle \sin 40^\circ = \frac{BF}{2}\)

\(\displaystyle BF \approx 1.29 miles\)

Fire tower B is \(\displaystyle 1.53 - 1.29 = 0.24\) miles closer to the fire.

In an angle-side-angle problem, Law of Sines will solve the triangle. 

First find angle A:

\(\displaystyle 180-(83+44)=53\)

Then use Law of Sines.

\(\displaystyle \frac {AB}{\sin 83^\circ} = \frac {4}{\sin 53^\circ} \rightarrow AB = \frac {4 \sin 83^\circ}{\sin 53^\circ} \approx 5.0\)