prove sin2x+cos2x+1 tan(pi/4+x)
______________ = __________
sin2x+cos2x-1 tanx
so what I did
tan(x) (1+cos(2x)+sin(2x)=tan(pi/4+x) (-1+cos(2x)+sin(2x)
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?
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.
To prove the given equation, we will simplify the expression on both sides and show that they are equal.
Starting with the left-hand side (LHS):
LHS = (sin^2(x) + cos^2(x) + 1) / (sin^2(x) + cos^2(x) - 1)
Recall the trigonometric identity: sin^2(x) + cos^2(x) = 1. Substituting this identity into the equation, we get:
LHS = (1 + 1) / (1 - 1)
= 2/0
Here, we encounter an issue because division by zero is undefined. Therefore, the LHS is not defined or does not exist.
Now, let's simplify the right-hand side (RHS):
RHS = tan(pi/4 + x) / tan(x)
Using the angle addition formula for tangent, we can rewrite tan(pi/4 + x) as:
tan(pi/4 + x) = (tan(pi/4) + tan(x)) / (1 - tan(pi/4)tan(x))
The value of tan(pi/4) is 1, so we have:
RHS = (1 + tan(x)) / (1 - tan(x))
Now, let's combine both sides of the equation:
LHS = RHS
2/0 = (1 + tan(x)) / (1 - tan(x))
Since division by zero is undefined, the equation does not hold for any value of x. We can conclude that the given equation is not true for all x.