Exact value of tan (x)

**tradar** · December 16th 2008, 05:11 AM

If sec (x) = -4/3 and sin (x) > 0, find the exact value of tan (x)...

**Grandad** · December 16th 2008, 05:21 AM

Hi tradar -

Originally Posted by tradar

If sec (x) = -4/3 and sin (x) > 0, find the exact value of tan (x)...

First you need to use:

$\sec^2 x = 1 + \tan^2 x$

This will give you a value for $\tan^2 x$ . When you take the square root, be careful with the $\pm$ sign. Which do you choose?

Well, we're told that $\sin x > 0$ and $\sec x < 0$ . Now $\cos x = \frac{1}{\sec x}$ . So $\cos x < 0$ . And, of course, $\tan x = \frac{\sin x}{\cos x}$ , so $\tan x < 0$ also.

So take the negative square root, and there's your answer.

Grandad

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