Prove that; tan20+4sin20= square root of3

tan( 20 ) + 4 Sin( 20 ) =

( Sin( 20 ) / Cos( 20 ) ) + 4 Sin( 20 ) =

Sin( 20 ) + 4 Sin( 20 ).Cos( 20 ) / Cos( 20 ) =

Sin( 20 ) + 2 × 2 Sin(20).Cos(20)/ Cos(20) =

Sin( 20 ) + 2 × Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Sin( 40 ) / Cos( 20 ) =

Sin( 20 ) + 2Cos( 50 ) / Cos ( 20 ) =

Sin( 20 ) + 2Cos( 20 + 30 ) / Cos( 20 ) =

2 × Cos( 30 + 20 ) =

2 × [ Cos(30).Cos(20) - Sin(30).Sin(20) ] =

2 × [ √3/2 × Cos(20) - 1/2 × Sin(20) ] =

√3 Cos(20) - Sin(20)

Sin( 20 ) + 2Cos ( 20 + 30 ) / Cos( 20 ) =

Sin( 20 ) + √3 Cos(20) - Sin(20) / Cos(20) =

Sin(20) - Sin(20) + √3 Cos(20) / Cos(20) =

0 + √3 Cos(20) / Cos(20) =

√3 Cos(20) / Cos(20) =

Cos(20) simplifies from the numerator and denominator of fraction

√3 × 1 / 1 =

√3

And we're done

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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'.