Question

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

Answer 1

-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^@13`
sin x = sin 8^@13 = 0.14 --> `csc x = 1/0.14 = 7.07`

b. tan y = 5 --> arc `y = 78^@69`
sin y = sin 78^@69 = 0.98
`csc y = 1/(sin 78^@69) = 1/0.98 = 1.02`
Finally,
S = - 707 + 2(1.02) = -5.03

Answer 2

`-sqrt50+(2sqrt26)/5approx-5.03146`

Explanation:

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

For `-csc("arccot"(7))`, 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 `"adjacent"=7` and `"opposite"=1`.

Through the Pythagorean Theorem, `"hypotenuse"=sqrt50`.

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

`csc("arccot"(7))=sqrt50/1`

and

`-csc("arccot"(7))=-sqrt50`

We can find `csc(arctan(5))` through a similar method.

If the tangent of an angle is `5`, then `"opposite"=5` and `"adjacent"=1`. Again, through the Pythagorean Theorem, we see that `"hypotenuse"=sqrt26`.

We want to find the cosecant of this angle as well, which will be `"hypotenuse"/"opposite"=sqrt26/5`.

Thus

`csc(arctan(5))=sqrt26/5`

and

`2csc(arctan(5))=(2sqrt26)/5`

So, combining these, we see that

`-csc("arccot"(7))+2csc(arctan(5))=-sqrt50+(2sqrt26)/5`