(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
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I don't understand the last part of the problem
Thank you. May I know how (sinθ/2 = cos θ/2)^2 = 1+sin θ ?this part I m not understanding.
Hi, I got it. Thank you very much for ur help! =D
Alternatively we can also get the solution as indicated
SORRY, I can't get the steps at line 2. sec^2 x -tan
SORRY, I can't get the steps at line 2. sec^2 x -tan^2 x -----> 1-(sec x -tan x) ? TQ.
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
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