Question

Help me please~
Prove that (1+cosθ+sinθ)/(1+cosθ-sinθ)=(1/cosθ)+tanθ

Answer 1

U can do it by taking conjugate pair.

Answer 2

How to do it?

Answer 3

Let x = 1+cosθ and y = sinθ

Answer 4

What is the next step?

Answer 5

There's lots of ways to deal with this and you're not going to be able to avoid some algebra here. So perhaps the easiest thing to do is simply to multiply it out:

I.e., expand the RHS (right hand side) and show it is equal to the LHS:

(1+cosθ+sinθ) = (1+cosθ-sinθ) [ (1/cosθ)+tanθ ]

Now,
RHS = (1+cosθ-sinθ) [ (1/cosθ)+tanθ ]
= 1/cosθ + tanθ + 1 + sinθ - tanθ - sinθ.tanθ
= 1/cosθ * ( 1 - sin^2 θ) + sinθ + 1
= cos^2 θ / cos θ + sin θ + 1
= cosθ + sinθ + 1
= LHS

Now I've left out what you probably think of some steps. So fill in this argument as you need to.

Answer 6

ok - this involves a lot of algebra im writing s for sin theta, c for cos theta and t for tan theta

so we have
(1 + c +s) / (1 + c -s)
using long division this gives 1 + 2s/(1 + c -s)

1 /c + t = = 1/c + s/c = (1+s)/c

now we need to show that 1 + 2s/(1 + c -s) is identical to (1 + s) / c

multiply through by lcm (c(1 + c + s):

(1+s) (1 + c - s) = = c + c^2 -cs + 2cs

lhs = 1 + c - s + s +sc - s^2 = 1 +c + sc - s^2..........(1)

rhs = c + c^2 + sc

now c^2 = 1 - s^2

so c + c^2 + sc = 1 + c + sc - s^2 which is eual to (1) above

that completes the proof

there might be an easier of doing this but this way is valid

Answer 7

Can I do it like this?

Answer 8

(1+cosθ+sinθ)/(1+cosθ-sinθ)=(1/cosθ)+tanθ

Answer 9

(1+cosθ+sinθ)/(1+cosθ-sinθ)=(1+sinθ)/cosθ

Answer 10

cosθ[(1+cosθ+sinθ)/(1+cosθ-sinθ)]=1+sinθ

Answer 11

cosθ+cos^2 θ+sinθcosθ=1+cosθ-sinθ+sinθ+cosθsinθ-sin^2 θ

Answer 12

cos^2 θ=1-sin^2 θ

Answer 13

cos^2 θ+sin^2 θ=1

Answer 14

1=1

Answer 15

∴(1+cosθ+sinθ)/(1+cosθ-sinθ)=(1/cosθ)+tan

Answer 16

Sure

Answer 17

And the fact you've found your own version of the proof is best of all.

Answer 18

Thank you.

Answer 19

^.^