Help with Proving Trig Identity

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nj_angel

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I have tried a couple of things and I can't seem to make any headway, the problem is:

(1+cos x+sin x)/ (1+cos x-sin x)= Sec x+Tan x

I have to prove how the left is the same as the right
 
nj_angel said:
I have tried a couple of things and I can't seem to make any headway
Click to expand...
What have you tried? (You may be almost there!)

nj_angel said:
(1+cos x+sin x)/ (1+cos x-sin x)= Sec x+Tan x
Click to expand...
If one groups explicitly, one can do things such as the following:

. . . . .(1 + [cos(x) + sin(x)]) / (1 + [cos(x) - sin(x)])

. . . . .{ (1 + [cos(x) + sin(x)]) / (1 + [cos(x) - sin(x)]) } { (1 - [cos(x) - sin(x)]) / (1 - [cos(x) - sin(x)]) }

. . . . .{ (1 + [cos(x) + sin(x)]) (1 - [cos(x) - sin(x)]) } / { 1 - [cos(x) - sin(x)]^2 }

. . . . .{ 1 - [cos(x) - sin(x)] + [cos(x) + sin(x)] - [cos(x) + sin(x)][cos(x) - sin(x)] } / { 1 - (cos^2(x) - 2sin(x)cos(x) + sin^2(x)) }

. . . . .{ 1 + 2sin(x) - [cos^2(x) - sin^2(x)] } / { 1 - (1 - 2sin(x)cos(x)) }

. . . . .{ 1 + 2sin(x) - cos^2(x) + sin^2(x) } / { 2sin(x)cos(x) }

Does this lead anywhere useful? Or would a different grouping perhaps work better? Or maybe starting from the right-hand side, instead? ;)
 
But wouldn't that way lead to 1/cosx? which is secx but it doesn't explain the tanx
 
nj_angel said:
I have tried a couple of things and I can't seem to make any headway, the problem is:

(1+cos x+sin x)/ (1+cos x-sin x)= Sec x+Tan x

I have to prove how the left is the same as the right
Click to expand...

When I have had to do things like prove an identity, I have generally started with the given identity and 'worked backwards'. For eample, assume the given formula is an identity and multiply by cos to clear the sec denominator, then by 1 + cos - sin to clear the denominator of the other side [now that we have only sines and cosines I'll just use s and c's]:
c (1 + c + s) = (1 + c - s) + s (1 + c - s)
and carry on from there eventually reaching
0 = 0.
Now work backwards again.
 
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