Find cosθ where tanθ=−n such that 0<n<1 and θ terminates in quadrant II.

Question

i am so confused please help

Answer 1

Make a figure!

The angle you want will be in quadrant 2.

The hypotenuse of the right triangle you need will be $\sqrt{(}{n}^2+1)$ (This intended to have the expression in parentheses ALL under the radical!)

The cos (theta) will be $\frac{-1}{\sqrt{(}{n}^2+1)}$ (again the whole parenthesis is under the radical!)