# Trigonometric Identities

• Nov 17th 2010, 11:33 PM
Ilsa
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.
• Nov 17th 2010, 11:55 PM
DrSteve
Multiply both sides of the equation by $\sin A$ to get $1 + \cos A = 3\sin A$.

Now square both sides to get $1 + 2\cos A + \cos^2 A = 9\sin^2 A = 9(1 - \cos^2 A)$

So $10\cos^2 A + 2\cos A - 8 = 0$.

$5\cos^ A + \cos A - 4 = 0$

$(5\cos A - 4)(\cos A + 1) = 0$

We get the two solutions $\cos A = 4/5, \cos A = -1$

-1 is an extraneous solution since it makes cot A undefined.
• Nov 18th 2010, 12:09 AM
Unknown008
Use what you were given:

$\csc A + \cot A = 3$

$1 + \cos A = 3 \sin A$ .....1

$\csc A - \cot A = \dfrac13$

$1 - \cos A = \dfrac{\sin A}{3}$

$\sin A = 3 - 3\cos A$ .........2

Substitute 2 in 1.

$1 + \cos A = 3(3 - 3\cos A)$

$1 + \cos A = 9-9\cos A$

$10 \cos A = 8$

$\cos A = \dfrac45$