How do you prove: #1- (sin^2x/(1-cosx))=-cosx#?

2 Answers
Feb 9, 2016

see explanation

Explanation:

To prove , require to manipulate one of the sides into the form of the other.
choosing the left side (LHS) gives

# 1 -(sin^2x/(1-cosx))#

require to combine these : rewrite # 1 = (1-cosx)/(1-cosx) #

now have : # (1-cosx)/(1-cosx) - sin^2x/(1-cosx) #

basically subtracting 2 fractions with a common denominator

hence #( 1-cosx-sin^2x)/(1-cosx) = ((1-sin^2x) -cosx)/(1-cosx)#

now # sin^2x + cos^2x = 1 rArr cos^2 x = 1 -sin^2x#

replacing this result into numerator

to obtain : #( cos^2x - cosx)/(1 - cosx) #

'taking' out a common factor of -cosx

# =( -cosx( 1 - cosx))/(1-cosx)#

# rArr -cosxcancel(1-cosx)/cancel(1-cosx) = - cosx #= RHS

thus proved

Feb 9, 2016

Alternate solution.

Explanation:

Left hand side (LHS) is

# 1 -(sin^2x/(1-cosx))#

We know the identity: #sin^2x+cos^2x=1#

#orsin^2x=1-cos^2x#

Substituting in the numerator of second term
# 1 -((1-cos^2x)/(1-cosx))#

Using the factors #x^2-y^2=(x+y)(x-y)#

We obtain
# 1 -((cancel((1-cosx))(1+cosx))/cancel(1-cosx))#

Simplifying we obtain
#1-(1+cosx)#
#or cancel1-cancel1-cosx#

# = - cosx = #RHS##

Hence proved