(1+cosx)/sinx + sinx/(1+cosx)
=
(1+cosx)^2 + sin^2x
------------------------
sinx(1+cosx)
1+2cosx+cos^2x+sin^2x
-------------------------
sinx(1+cosx)
= (2+2cosx)/(sinx(1+cosx))
= 2/sinx
so, does
2/sinx = 1
have any solutions?
Solve for:
1+cosx/sinx + sinx/1+cosx = 1
My teacher said no solution, but I don't know how he got that.
3 answers
cos x/sin x + sin x/(1+cos x) = 0 ??
nah, maybe you mean
(1+cos x)/sin x + sin x/(1+cos x)=1
(1+cos)(1-cos)/(sin(1-cos))
+sin (1-cos)/[(1+cos)(1-cos)] = 1
(1-cos^2)/(sin-sin cos)
+ (sin-sin cos)/(1-cos^2) = 1
sin/(1-cos) + (1-cos)/sin = 1
sin^2 + (1-cos)^2 = sin(1-cos)
(1-cos^2) +1 - 2 cos + cos^2 = sin-sin cos
2 - 2 cos = sin (1-cos)
humm
2 ( 1-cos) = sin(1-cos)
sin x = 2 !!!!!
well
-1 </= sin x </= +1
so impossible
nah, maybe you mean
(1+cos x)/sin x + sin x/(1+cos x)=1
(1+cos)(1-cos)/(sin(1-cos))
+sin (1-cos)/[(1+cos)(1-cos)] = 1
(1-cos^2)/(sin-sin cos)
+ (sin-sin cos)/(1-cos^2) = 1
sin/(1-cos) + (1-cos)/sin = 1
sin^2 + (1-cos)^2 = sin(1-cos)
(1-cos^2) +1 - 2 cos + cos^2 = sin-sin cos
2 - 2 cos = sin (1-cos)
humm
2 ( 1-cos) = sin(1-cos)
sin x = 2 !!!!!
well
-1 </= sin x </= +1
so impossible
Do I get extra credit for finding a harder way ?