# Prove that sinB/1-cosB -1+cosB/sinB =0

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### 4 Answers

sinB/(1-cosB) - (1+cosB)/sinB =0

L.H.S= sinB/(1-cosB) - (1+cosB)/sinB

={2sin(B/2)cos(B/2)}/[1-1+2sin^2(B/2)] - [1+2cos^2(B/2)-1]/{2sin(B/2)cos(B/2)}

=cot(B/2)-cot(B/2)

=0

=R.H.S

We have to prove:

[sinB/(1 - cosB)] - [(1 + Cos B)/sinB] = 0

To do this we multiply both sides of the equation by (1 - cosB)*sinB. This gives us:

[sinB/(1 - cosB)]*[(1 - cosB)*sinB] - [(1 + Cos B)/sinB][(1 - cosB)*sinB] = 0*[(1 - cosB)*sinB]

Simplifying this we get:

[sinB*sinB] - [(1 + Cos B)*(1 - cosB)] = 0

Or

sinB^2 - 1 + cosB^2 = 0

Shifting the term -1 on the left hand side of the above equation of the right hand side we get:

sinB^2 + cosB^2 = 1

We know the above equation to be true. There the original equation we stared with is also true.

We have to edit the given expression as below:

sinB/(1-cosB) - (1+cosB)/sinB = 0 for the expression to be an identity for the reason given below:

Please use the bracket as otherwise , the given expression, sinB/1-cosB/sinB = 0 is equivalent to (sinB/1)-cosB+(cosB/sinB) = 0 which is not an identity , as for B =0 degree, 0/1-1+1/0 is ifinity. So the given expression, as it is without editting, is disproved tobe not an identity.

Now, to prove that

sinB/(1-cosB) -(1+cosB)/sinB = 0

Proof:

We know that sin^2 B+ cos^2 B = 1 or

sin^2B = 1-cos^2B =(1+cosB)(1-cosB)

SinB * sinB = (1+cosB)(1-cosB) or

SinB/(1-cosB) = (1+cosB)/sinB or

sinB/(1-cosB) -(1+cosB)/sinB = 0

To make the subtraction between the two fractions, they have to have the same denominator. For this reason, we'll do the amplification of the first fraction, with the denominator of the second, namely sinB, and we'll do the amplification of the second fraction with the denominator of the first one, namely (1-cosB).

The expression will appear in this way:

[(sinB*sinB)-(1-cosB)*(1+cosB)] : [sinB*(1-cosB)]=0

Also, doing the cross muliplying between the left terms of the expression and the right terms, we'll write, even more simple:

[(sinB*sinB)-(1-cosB)*(1+cosB)]=[sinB*(1-cosB)]*0

[(sinB*sinB)-(1-cosB)*(1+cosB)]=

Now, all we have to do is to open the brackets and solve the expression:

sinB*sinB= (sinB)^2

(1-cosB)*(1+cosB)=1-(cos B)^2

(sinB)^2-[1-(cos B)^2]=0

(sinB)^2-1+(cos B)^2=0

We note that in the expression above, are added the squares of the joint functions, (SIN B)^2 and (COS B)^2,of the same angle B.

(sinB)^2 + (cos B)^2=1

We'll substitute the adding with sum, in the expression above:

(sinB)^2-1+(cos B)^2=0 => 1-1=0 => **0=0 TRUE**