sin x/1 + cos x + 1 + cos x/sin x = 2 csc x
Show that the left side equals the right side.
I think this is supposed to be :
sin x / (1 + cos x) + (1 + cos x)/sin x = 2 csc x
get a common denominator on the left = sinx (1 + cos x)
[ sinx (sinx) + ( 1 + cosx)^2 ] / [ sinx (1 + cosx)]
[sin^2x + 1 + 2cosx + cos^2x] / [ sinx(1 + cosx)]
[sin^2x + cos^2x + 1 + 2cosx] / [ sinx(1 + cosx)]
{remember : sin^2x + cos^2x = 1}
[ 1 + 1 + 2 cosx] / [sinx (1 + cosx)]
[2 + 2cos x] / [ sinx (1 + cos x) ] factor out 2 on top
2(1 + cosx) / [sinx ( 1 + cosx)] "cancel " the (1 + cos x) on top/bottom
2 / sinx =
2 (1 / sinx) =
2csc x and this = the right hand side
I think this is supposed to be :
sin x / (1 + cos x) + (1 + cos x)/sin x = 2 csc x
get a common denominator on the left = sinx (1 + cos x)
[ sinx (sinx) + ( 1 + cosx)^2 ] / [ sinx (1 + cosx)]
[sin^2x + 1 + 2cosx + cos^2x] / [ sinx(1 + cosx)]
[sin^2x + cos^2x + 1 + 2cosx] / [ sinx(1 + cosx)]
{remember : sin^2x + cos^2x = 1}
[ 1 + 1 + 2 cosx] / [sinx (1 + cosx)]
[2 + 2cos x] / [ sinx (1 + cos x) ] factor out 2 on top
2(1 + cosx) / [sinx ( 1 + cosx)] "cancel " the (1 + cos x) on top/bottom
2 / sinx =
2 (1 / sinx) =
2csc x and this = the right hand side