I need help with basic trig. identities.?

Question

prove that 1-cos(x)/sin(x) = sin(x)/1+cos(x)

Answer 1

Multiplying numenator & denominator of the L.H.S by:
(1+cos(x))
L.H.S = [(1-cos(x))(1+cos(x))] / [sin(x)(1+cos(x))]
= [ 1- cos^2(x)]/ [sin(x)(1+cos(x))]
= [sin^2(x)]// [sin(x)(1+cos(x))]
= sin(x)/(1+cos(x)) = R.H.S

Answer 2

sin^2(x) + Cos^2(x) =1

so, you have:
1-cos(x)/ Sin(x) = sin(x)/ (1+cos(x))

form there:

(1- Cos^2(x)) = Sin^2(x)

and then
1 = Sin^2(x) + Cos^2(x)

and there is the identity

Answer 3

multiply both sides of the equation by (sin(x))(1+cos(x))
and you get
1-cos^2(x) = sin^2(x)
1 = cos^2(x) + sin^2(x)
That's it.

Answer 4

LHS
= [(1 - cos x) / sinx ] / [ (1 + cos x) / (1 + cos x)]
= (1 - cos² x ) / ( sinx.(1 + cos x) )
= sin² x / ( sinx.(1 + cos x) )
= sin x / (1 + cosx)
= RHS

Answer 5

The problem to solve is:
sin(x)/1+cos(x)


sin(x)/1 evaluates to sin(x)

sin(x)/1+cos(x) evaluates to sin(x)+cos(x)



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The final answer is sin(x)+cos(x)

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Now, Webmath will try to simplify this using trigonometric identities

Answer 6

RHS
= sinx / (1+cosx)
= [sinx / (1+cosx)][(1-cosx) / (1-cosx)]
= [sinx(1-cosx)] / [(1+cosx)(1-cosx)]
= [sinx(1-cosx)] / [1-(cosx)^2]
= [sinx(1-cosx)] / (sinx)^2
= (1-cosx) / sinx
= LHS (QED)