Express Tan x as function of sin only

Tan x = sin x / √[1 - sin²θ]

Step-by-step explanation:

We know that;

Tanx = Sinx / cosx

Squaring both side

Tan²x = Sin²x / cos²x

[cos²θ = 1 - sin²θ]

So,

Tan²x = Sin²x / [1 - sin²θ]

Taking root both side

Tan x = sin x / √[1 - sin²θ]

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.