# Prove!

• Sep 5th 2013, 08:12 AM
Shayna
Prove!
Hi,

Prove that:
(sin θ-kos θ+1)/(sin θ+kos θ-1)= (sin θ+1)/(kos θ)

(kos θ/ 1-tan x) + (sin θ/1-kot θ) = sin θ + kos θ

Solve that tanx = π - 2x , 0<x<π/2

Thnks. :D
• Sep 5th 2013, 08:44 AM
red_dog
Re: Prove!
1) $\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2} +2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}=\frac{\sin\frac{\theta}{2 }+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=\\=\frac{\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}=\frac{1+\sin \theta}{\cos \theta}$

2) $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\frac{\cos^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin^2 \theta}{\sin \theta-\cos \theta}=\\=\frac{\cos^2 \theta-\sin^2 \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta$

I don't understand the last part of the problem
• Sep 5th 2013, 10:10 AM
Shayna
Re: Prove!
Thank you. May I know how (sinθ/2 = cos θ/2)^2 = 1+sin θ ?this part I m not understanding.
• Sep 5th 2013, 10:26 AM
Shayna
Re: Prove!
Hi, I got it. Thank you very much for ur help! =D
• Sep 5th 2013, 09:46 PM
ibdutt
Re: Prove!
Alternatively we can also get the solution as indicated
Attachment 29123
• Sep 6th 2013, 01:26 AM
Shayna
Re: Prove!
hi,

SORRY, I can't get the steps at line 2. sec^2 x -tan
• Sep 6th 2013, 01:27 AM
Shayna
Re: Prove!
hi,

SORRY, I can't get the steps at line 2. sec^2 x -tan^2 x -----> 1-(sec x -tan x) ? TQ.
• Sep 6th 2013, 02:59 AM
ibdutt
Re: Prove!
It was (sec x +tan x ) -1 = ( sec x - tan x ) - ( sec^2 x - tan^2 x ) .... Because sec^2 x -tan^2 x = 1
= ( sec x - tan x ) - (( sec x - tan x ) ( sec x +tan x ) )= ( sec x +tan x ) [ 1 - ( sec x - tan x ) ] etc