solve this equation for A in the range 0<A<360

tanA + cotA = 2secA

i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this?

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- Jul 5th 2009, 11:26 PM #1

- Jul 5th 2009, 11:39 PM #2
$\displaystyle \tan{A} + \cot{A} = 2\sec{A}$

$\displaystyle \frac{\sin{A}}{\cos{A}}+ \frac{\cos{A}}{\sin{A}} = \frac{2}{\cos{A}}$

$\displaystyle \frac{\sin^2{A}}{\sin{A}\cos{A}} + \frac{\cos^2{A}}{\sin{A}\cos{A}} = \frac{2\sin{A}}{\sin{A}\cos{A}}$

$\displaystyle \frac{\sin^2{A} + \cos^2{A}}{\sin{A}\cos{A}} = \frac{2\sin{A}}{\sin{A}\cos{A}}$

$\displaystyle \sin^2{A} + \cos^2{A} = 2\sin{A}$

$\displaystyle 1 = 2\sin{A}$

$\displaystyle \sin{A} = \frac{1}{2}$

Since $\displaystyle \sin$ is only positive in the first and second quadrants, in the domain $\displaystyle 0^{\circ} \leq x \leq 360^{\circ}$, there will only be two solutions.

$\displaystyle A = 30^{\circ}$ or $\displaystyle A = 150^{\circ}$.

- Jul 6th 2009, 12:34 AM #3
Hello furor celticaProveIt has given you the correct solution. But to try to answer your question, did you, in your solution, square both sides of the equation at some point? If you did, that will be why you got the extra, unwanted solutions.

You can see a similar thing if you square both sides in a simple non-trig equation. For example:

$\displaystyle x+1=4$

The solution is, of course, $\displaystyle x = 3$. But if you square both sides:

$\displaystyle (x+1)^2 = 16$

$\displaystyle \Rightarrow x^2 + 2x + 1 = 16$

$\displaystyle \Rightarrow x^2 +2x - 15=0$

$\displaystyle \Rightarrow (x +5)(x-3)=0$

$\displaystyle \Rightarrow x = -5$ or $\displaystyle 3$

And the value $\displaystyle x = -5$ doesn't satisfy $\displaystyle x+1 = 4$, but $\displaystyle x+1 = -4$. But of couse, both $\displaystyle 4^2$ and $\displaystyle (-4)^2$ give the value $\displaystyle 16$.

So, always beware of squaring both sides of an equation: you'll sometimes introduce extra, unwanted solutions if you do.

Grandad

- Jul 6th 2009, 03:31 AM #4