Prove that #(sint-cost)/(sint+cost) = (2sin^2t-1)/(1+2sintcost)#?

2 Answers
Apr 5, 2018

Well, 1 plays an important role in solving this problem.

Explanation:

Let us focus on the Left Hand Side (LHS)

#LHS=(sint -cost)/(sint+cost)#

Note that #1 = (sint +cost)/(sint +cost)#

We will use this property to aid us because multiplying our LHS by 1 does not change its value or solutions.

That is,
#LHS=(sint -cost)/(sint+cost)*(sint +cost)/(sint +cost #
Now we simplify, the numerator is a difference of two squares. Expand the numerator and rearranging terms, we have:

#LHS= (sin^2t-cos^2t)/(sint + cost)^2#

#LHS= (2sin^2t - 1)/(sin^2t +cos^2t+2sintcost)#

We know that #sin^2t+cos^2t=1#

So we have:

#LHS=(2sin^2t - 1)/(1+2sintcost)#

This is equal to the RHS. QED

Apr 5, 2018

Please see below.

Explanation:

#(sint-cost)/(sint+cost)#

= #(sint-cost)/(sint+cost)*(sint+cost)/(sint+cost)#

= #(sin^2t-cos^2t)/(sin^2t+cos^2t+2sintcost)#

= #(sin^2t-(1-sin^2t))/(1+2sintcost)#

= #(sin^2t-1+sin^2t)/(1+2sintcost)#

= #(2sin^2t-1)/(1+2sintcost)#