((10分題))簡單數學公式!!(急趕)

(a)

Left Side
tanx sinx + cosx
[ tan x = sin x/cos x ]
= sin^2 x/cosx + cosx
= ( sin^2 x + cos^2 x)/cos x
[sin^2 x + cos^2 x = 1]
= 1/cos x

Right Side
1/cos x

Left side = Right Side

(b) [sin^2 x + cos^2 x = 1]

Left Side
sin^4 x - cos^4x
= 1-cos^4x - cos^4x
=1

Right Side
sin^2 x - cos^2 x
= 1 - cos^2 x - cos^2 x
= 1

Left Side = Right Side

As follow AS~~~~

圖片參考：http://g.imagehost.org/0784/ScreenHunter_03_Feb_22_11_44.gif

a)LHS
=tanx sinx +cosx
=(sinx/cosx)*sinx +cosx
=(sin^2 x /cosx)+cosx
=(sin^2x+cos^2 x)/cosx
=1/cosx
=RHS//

b)
LHS
=sin^4 x-cos^4x
=(sin^2x )^2 -(cos^2 x)^2
=(sin^2x -cos^2x)(sin^2x+cos^2x)
=sin^2x-cos^2x
=RHS//

tan(x) = sin(x)/cos(x) and
sin(x)^2 + cos(x)^2 = 1

tan(x)sin(x) + cos(x)
= (sin(x)^2/cos(x) + cos(x)^2/cos(x))
= 1/cos(x)

sin(x)^4 - cos(x)^4
= (sin(x)^2 + cos(x)^2)(sin(x)^2 - cos(x)^2)
= sin(x)^2 - cos(x)^2