prove sin2x+cos2x+1 tan(pi/4+x)

Question

______________ = __________
sin2x+cos2x-1 tanx

so what I did

tan(x) (1+cos(2x)+sin(2x)=tan(pi/4+x) (-1+cos(2x)+sin(2x)

Answer 1

sin2x+cos2x
= √2(sin2x * 1/√2 + cos2x * 1/√2)
= √2sin(2x+π/4)
so you have

[√2sin(2x+π/4)+1]/[√2sin(2x+π/4)-1]
= [√2sin(2x+π/4)+1]^2/(2sin^2(2x+π/4)-1)

Now, using sum-to-product formulas,

√2sinu+1 = √2(sinu + 1/√2)
= √2(sinu + sin π/4)
= √2*2sin((u+π/4)/2)cos((u-π/4)/2)
= 2√2sin(x+π/4)cos(x)
similarly,
√2sinu-1 = 2√2cos(x+π/4)sin(x)

So, the left side is now

2√2sin(x+π/4)cos(x)
-----------------------
2√2cos(x+π/4)sin(x)

I think you can now see your way clear, right?

Answer 2

skip that middle section. It was a blind alley I forgot to erase.

Cool problem. I had to try several things before I hit on the right trick to convert all the sums to products.