How do you simplify #cotB + (sinB / (1+cosB))#?

Question

4789 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-05T06:34:38

#cscbeta#

#cotbeta#+#sinbeta/(1+cosbeta)#

1) Turn everything into sin and cos.

#cosbeta/sinbeta+sinbeta/(1+cosbeta)#

2) Use LCD and create a common numerator.

#(1+cosbeta)/(1+cosbeta)*cosbeta/sinbeta+sinbeta/(1+cosbeta)*sinbeta/sinbeta#

#((1+cosbeta)cosbeta+sin^2beta)/(sinbeta(1+cosbeta))#

3) Distibute #cosbeta# in the denominator.

#(cosbeta+(cos^2beta+sin^2beta))/(sinbeta(1+cosbeta))#

4) Pythagorean Identity in the denominator. Then cancel out matching denominator and numerator.

# cancel(cosbeta+(1))/(sinbeta (cancel(1+cosbeta)))#

5) Left with #1/sinbeta# = #cscbeta#

Answer 2 · 2015-11-05T06:38:30

#csc(B)#

#cot(B)+frac{sin(B)}{1+cos(B)}=frac{cos(B)}{sin(B)}+frac{sin(B)}{1+cos(B)}frac{1-cos(B)}{1-cos(B)}#

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

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

#=frac{cos(B)}{sin(B)}+frac{1-cos(B)}{sin(B)}#

#=frac{1}{sin(B)}#

#=csc(B)#