# Prove LHS is equal to the RHS?

( sin (A) / ( 1 + cos (A) ) ) + ( (1 + cos (A) ) / sin (A) ) = 2 / sin (A)

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• Anonymous
9 years ago

Start out by getting common denominators and adding the fractions to get:

LHS = sin(A)/[1 + cos(A)] + [1 + cos(A)]/sin(A)

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

= {sin^2(A) + [1 + cos(A)]^2}/{sin(A)[1 + cos(A)]}, by adding numerators

= [sin^2(A) + cos^2(A) + 2cos(A) + 1]/{sin(A)[1 + cos(A)]}, by expanding

= [2cos(A) + 2]/{sin(A)[1 + cos(A)]}, since sin^2(A) + cos^2(A) = 1

= {2[1 + cos(A)]}/{sin(A)[1 + cos(A)]}, by factoring out the 2

= 2/sin(A), by canceling 1 + cos(A)

= RHS.

I hope this helps!

• Anonymous
9 years ago

LHS = ( sin (A) / ( 1 + cos (A) ) ) + ( (1 + cos (A) ) / sin (A) )

= {sin^2(A) + cos^2(A) + 2cos(A) + 1}/[sin(A)*(1+cos(A)]

= 2[(1+cos(A)]/[sin(A)*(1+cos(A)]

= 2/sin(A) = RHS. ///

• 9 years ago

2/sin(a) = 2/sin(a)

there i just did haha

I would answer this question if we were talking one-on-one, solving online is hard.

But let me give you an advice, just in case you were ask to solve such thing in your exams/tests, do what I did, just copy one side to the other, you'll get at least a mark for that.

• 9 years ago

(sinA)/(1+cosA) = [(sinA)^2]/[(sinA)(1+cosA) = [1-(c0sA)^2]/(sinA)(1+cosA)=(1-cosA)/sinA

now write LHS with this expression and add the 2 fractions to get the RHS.