If sec (x) = -4/3 and sin (x) > 0, find the exact value of tan (x)...
Hi tradar -
First you need to use:
$\displaystyle \sec^2 x = 1 + \tan^2 x$
This will give you a value for $\displaystyle \tan^2 x$. When you take the square root, be careful with the $\displaystyle \pm$ sign. Which do you choose?
Well, we're told that $\displaystyle \sin x > 0$ and $\displaystyle \sec x < 0$. Now $\displaystyle \cos x = \frac{1}{\sec x}$. So $\displaystyle \cos x < 0$. And, of course, $\displaystyle \tan x = \frac{\sin x}{\cos x}$, so $\displaystyle \tan x < 0$ also.
So take the negative square root, and there's your answer.
Grandad