Find theta: Solving csc & ctn Equations with x-6 & x+5

[lucifer_x]

if

csc theta=
x-6
x-2

and

ctn theta=
x+5
x-2

find theta

 

[Integral]

You need to show us some work.

Here are some questions to get you moving.

What is sin $\theta$ ?

What is Tan $\theta$ ?

Now use what you know about the definition of sin and Tan to draw a triangle. Apply Pythagorean theorem.

 

[Defennder]

First you need to solve for x. That means making use of a trigo identity to get theta out of the way. Express sin theta and cos theta in terms of x, then apply that trigo identity.

 

[HallsofIvy]

Trig identities are one way to shift from one function to another (for example, if you know sec(x)= a/b and want to find tan(x), write a trig identity that converts sec to tan, say tan²(x)= sec²(x)- 1 and so that tan²(x)= x²- 1 and tan(x)= \sqrt{a^2/b^2- 1}= .\sqrt{a^2- b^2}/b [/itex].

Still another, perhaps simpler, method is to draw a triangle with one angle x and write in appropriate lengths for the sides. In the above example, if sec(x)= a/b then, since sec= 1/cos or "hypotenuse over near side", our triangle would have hypotenuse a and near side b. By the Pythagorean theorem, the "opposite side" has length $\sqrt{a^2- b^2}$ and so tan(x), "opposite side over near side", is $\sqrt{a^2- b^2}/b$

But don't use the word "reciprocals" here. Reciprocals are specifically the "multiplicative inverses". Here you are concerned with "functional inverses".