
Trigonometric Identities
Given cosec A + cot A = 3, evaluate cosec A  cot A and cos A.
I was able to evaluate cosec A  cot A:
cosec^2 A  cot^2 A = 1
(cosec A  cot A)(cosec A + cot A) = 1
(cosec A  cot A) x 3 = 1
Therefore, (cosec A  cot A) = 1/3
However, I was not able to evaluate cos A.
The answer should come up to cos A = 4/5.

Multiply both sides of the equation by $\displaystyle \sin A $ to get $\displaystyle 1 + \cos A = 3\sin A$.
Now square both sides to get $\displaystyle 1 + 2\cos A + \cos^2 A = 9\sin^2 A = 9(1  \cos^2 A)$
So $\displaystyle 10\cos^2 A + 2\cos A  8 = 0$.
$\displaystyle 5\cos^ A + \cos A  4 = 0$
$\displaystyle (5\cos A  4)(\cos A + 1) = 0$
We get the two solutions $\displaystyle \cos A = 4/5, \cos A = 1$
1 is an extraneous solution since it makes cot A undefined.

Use what you were given:
$\displaystyle \csc A + \cot A = 3$
$\displaystyle 1 + \cos A = 3 \sin A$ .....1
$\displaystyle \csc A  \cot A = \dfrac13$
$\displaystyle 1  \cos A = \dfrac{\sin A}{3}$
$\displaystyle \sin A = 3  3\cos A$ .........2
Substitute 2 in 1.
$\displaystyle 1 + \cos A = 3(3  3\cos A)$
$\displaystyle 1 + \cos A = 99\cos A$
$\displaystyle 10 \cos A = 8$
$\displaystyle \cos A = \dfrac45$