# What does -csc(arc cot(7))+2csc(arctan(5)) equal?

Jun 10, 2016

-5.03

#### Explanation:

S = -csc(arccot (7)) + 2csc(arctan (5))
Use calculator -->
a. cot x = 7 --> tan x = 1/7
arccot(7) = arctan(1/7) --> arc $x = {8}^{\circ} 13$
sin x = sin 8^@13 = 0.14 --> $\csc x = \frac{1}{0.14} = 7.07$

b. tan y = 5 --> arc $y = {78}^{\circ} 69$
sin y = sin 78^@69 = 0.98
$\csc y = \frac{1}{\sin {78}^{\circ} 69} = \frac{1}{0.98} = 1.02$
Finally,
S = - 707 + 2(1.02) = -5.03

Jun 12, 2016

$- \sqrt{50} + \frac{2 \sqrt{26}}{5} \approx - 5.03146$

#### Explanation:

This is solvable without a calculator. It all depends on drawing pictures of the triangles.

For $- \csc \left(\text{arccot} \left(7\right)\right)$, this is asking for the cosecant of the triangle where the cotangent is already equal to $7$.

Since cotangent is equal to the adjacent side of the angle in question divided by the opposite side, we can say that $\text{adjacent} = 7$ and $\text{opposite} = 1$.

Through the Pythagorean Theorem, $\text{hypotenuse} = \sqrt{50}$.

Since we want to find cosecant of this triangle, we will take the hypotenuse over the opposite side, so

$\csc \left(\text{arccot} \left(7\right)\right) = \frac{\sqrt{50}}{1}$

and

$- \csc \left(\text{arccot} \left(7\right)\right) = - \sqrt{50}$

We can find $\csc \left(\arctan \left(5\right)\right)$ through a similar method.

If the tangent of an angle is $5$, then $\text{opposite} = 5$ and $\text{adjacent} = 1$. Again, through the Pythagorean Theorem, we see that $\text{hypotenuse} = \sqrt{26}$.

We want to find the cosecant of this angle as well, which will be $\text{hypotenuse"/"opposite} = \frac{\sqrt{26}}{5}$.

Thus

$\csc \left(\arctan \left(5\right)\right) = \frac{\sqrt{26}}{5}$

and

$2 \csc \left(\arctan \left(5\right)\right) = \frac{2 \sqrt{26}}{5}$

So, combining these, we see that

$- \csc \left(\text{arccot} \left(7\right)\right) + 2 \csc \left(\arctan \left(5\right)\right) = - \sqrt{50} + \frac{2 \sqrt{26}}{5}$