If secx = -3, and x lies in quadrant II find tan x/2

-√2 or √3/3?

sec x = -3 ---> cosx = -1/3

sinx = 2√2/3 , in II the sine is + , and I used Pythagoras

tan x/2= sinx/2 / cos x/2

Now look how I answered your second question in
http://www.jiskha.com/display.cgi?id=1359587430

This one is very similar to it
give it a shot

btw, my answer was +√2

(since x was in the second quadrant, x/2 must be in the first, where everything is positive. So -√2 would be wrong without even doing any calculations)

Okay I think I sort of understand it now. Thank you so much, I really appreciate it

To find the value of tan(x/2), we can use the half-angle identities for the tangent function. However, before we can do that, we need to determine the value of x.

Given that sec(x) = -3 and x lies in quadrant II, we know that sec(x) is negative in quadrant II. Since sec(x) is the reciprocal of cosine, it means that cosine is negative in quadrant II.

Let's try to find the value of cosine (cos(x)) first.

Recall that the secant function is the reciprocal of cosine:

sec(x) = 1/cos(x)

Given sec(x) = -3, we can rewrite it as:

1/cos(x) = -3

Now, cross-multiply:

1 = -3cos(x)

Divide both sides by -3:

cos(x) = -1/3

Since cosine is negative in quadrant II, cos(x) = -1/3 satisfies this condition.

Now, let's use the half-angle identity for tangent:

tan(x/2) = ±√((1 - cos(x))/(1 + cos(x)))

Substituting the value of cos(x):

tan(x/2) = ±√((1 - (-1/3))/(1 + (-1/3)))

Simplifying:

tan(x/2) = ±√((4/3)/(2/3))

Dividing the fractions:

tan(x/2) = ±√(4/2)

Simplifying the square root:

tan(x/2) = ±√2

Therefore, the value of tan(x/2) when sec(x) = -3, and x lies in quadrant II, can be either -√2 or +√2.

Note: The other option you mentioned, √3/3, does not apply in this situation and is not the correct value.