how do i find cosX if i am given sin x= -4/5 and tan <o???

Isn't the sine negative in the III and IV quadrants, and isn't the tangent negative in the II and IV quadrant.

So clearly our angle x must be in IV

Make a triangle using sinx = 4/5
The third side (the adjacent) would be 3 and cosx = 3/5, but we are in quadrant IV, and in IV the cosine is positive, so ..
cosx = 3/5

cos^2x+ sin^2 x=1

the tangent being negative tells you which of the two answers is correct.

To find the value of cos(x), you can use the given information that sin(x) = -4/5 and tan(x) = <o.

First, recall that sin(x) is the ratio of the length of the side opposite angle x to the length of the hypotenuse in a right triangle. In this case, sin(x) = -4/5, which means that the side opposite angle x has a length of -4 and the hypotenuse has a length of 5.

Next, recall that tan(x) is the ratio of the length of the side opposite angle x to the length of the side adjacent to angle x in a right triangle. Since we know the side opposite angle x is -4, we need to find the length of the side adjacent to angle x. Let's call this length "a". Therefore, tan(x) = (-4)/a = <o.

Now, we can use the Pythagorean theorem to find the length of the side adjacent to angle x. The Pythagorean theorem states that in a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.

Applying the Pythagorean theorem, we have (-4)^2 + a^2 = 5^2. Simplifying this equation gives 16 + a^2 = 25.

Subtracting 16 from both sides gives a^2 = 9. Then, taking the square root of both sides gives a = 3, since we are dealing with a positive length.

Now that we know the lengths of the two sides, we can find cos(x). Recall that cos(x) is the ratio of the length of the side adjacent to angle x to the length of the hypotenuse in a right triangle. In this case, cos(x) = 3/5, since the length of the side adjacent to angle x is 3 and the length of the hypotenuse is 5.

Therefore, the value of cos(x) is 3/5.