Question

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

Answer 1

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

Answer 2

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)`