Given cosa = - 3 / 5, and a is the second quadrant angle, find Sina and Tana

Given cosa = - 3 / 5, and a is the second quadrant angle, find Sina and Tana

sina=4/5
tana= -4/3

Given that sina = - √ 3,2, and a is the fourth quadrant, find the value of cosa, Tana

A is the fourth quadrant
cossa>0
sin²a+cos²a=1
So cosa = 1 / 2
tana=sina/cosa=-√3

How to find the maximum and minimum of F (a) = sina-1 / cosa-2?

Let y = sina-1, x = cosa-2 (Sina) ^ 2 + (COSA) ^ 2 = 1 (x + 2) ^ 2 + (y + 1) ^ 2 = 1, then f (a) = Y / x, let Y / x = k, y = KX. The problem is that when a circle and a straight line have a common point, the value of K, then when the straight line is tangent, the center of the circle (- 2, - 1), the radius = 1, and the distance from the center of the circle to the tangent is equal to the radius kx-y = 0 | - 2K + 1 | / √ (k ^ 2

Find the minimum value of the square of cosa minus the square of sina

Formula of reducing times and formula
Reduction times:
sin2θ=(1-cos2θ)/2
cos2θ=(1+cos2θ)/2
Square of cosa = (1 + cos2a) / 2
Square of sina = (1-cos2a) / 2
The square of cosa minus the square of sina = (1 + cos2a) / 2 - (1-cos2a) / 2 = cos2a > = - 1
The minimum is - 1