2008/004) If α + β + γ = π/2, show that,

If α + β + γ = π/2, show that,

[(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan α/2)(1 + tan α/2)]
= (sin α + sin β + sin γ - 1)/(cos α + cos β + cos γ)

proof:
We should start with the RHS as it is more complex

Before we proceed let us use/deduce certain information as they shall be used
(actually they should be derived as required but I am deriving to keep the flow
α + β + γ = π/2
so α + β = π/2 – γ
or α + β = π/2 – γ …1

cos a + cos b = 2 cos (a+b)/2 cos(a-b)/2 … 2

taking sin of both sides of 1 we get
sin (α + β) = sin (π/2 – γ) = cos γ … 3
sin 2a = 2 sin a cos a …. 4

from 1
(α + β)/2 = (π/2 – γ)/2
So sin (α + β)/2 = sin (π/2 – γ)/2 = cos (π/2 + γ)/2 … 5

Again as
α + β + γ = π/2
so α - β + γ = π/2 - 2 β
so α - β + γ + π/2 = π - 2 β
so (α - β + γ + π/2)/ 4 = π/4 - β/2 …6

Again as
α + β + γ = π/2
so α - β - γ = π/2 - 2 β – 2 γ
so α - β - γ - π/2 = - 2 β – 2 γ
so (α - β - γ - π/2)/ 4 = - (β + γ)/2
so cos (α - β - γ - π/2)/ 4 = cos (β + γ)/2 as cos - A= cos A
= cos (π/2 – α)/2
= cos (π/4 – α/2) .. 7
And sin (α - β - γ - π/2)/ 4 = - sin (β + γ)/2 = - sin (π/4 – α/2)
Or sin (-α + β +γ + π/2)/ 4 = sin (π/4 – α/2) … 8


Now let us find he denominator
cos α + cos β + cos γ
= 2 cos (α + β)/2 cos(α - β )/2 + cos γ (using 2)
= 2 cos (α + β)/2 cos(α - β )/2 + sin (α + β) (using 3)
= 2 cos (α + β)/2 cos(α - β )/2 + 2 cos (α + β)/2 sin (α + β)/2 ( using 4)
= 2 cos (α + β)/2 (cos(α - β )/2 + sin (α + β)/2)
= 2 cos (π/2 - γ)/2 (cos(α - β )/2 + cos (π/2 + γ)/2)) (using 1 and 5)
= 2 cos (π/4 – γ/2)*2 cos (α - β + π/2 + γ)/2 cos (α - β - π/2 - γ)/2 (using 2)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (α - β - π/2 - γ)/2 (using 6)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) using 7

So cos α + cos β + cos γ = 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) … (A)
Now numerator

For the additional identities

Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2 … 9
sin (α + β + γ) = sin π/2 = 1 …. 10

sin A – sin B = 2 sin (A-B)/2 cos (A+B)/2 … 11


cos A – cos B =2 sin (A+B)/2 sin (B-A)/2 … 12

tan (π/4- A)= (1- tan π/4 tan A) / (tan π/4 + tan A) = ( 1- Tan A)/(1+tan A) … 13

based on above numerator

sin α + sin β + sin γ – 1
= 2 sin (α + β)/2 cos(α - β )/2 + sin γ – sin (α + β + γ) (using 9 and 10)
= 2 sin (π/4 – γ/2) cos(α - β )/2 –(sin (α + β + γ) - sin γ) (using 5)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (α + β)/2 cos (α + β + γ) + γ)/2 (using 11)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (π/4 – γ/2) cos (π/2+ γ)/2 (using 5)
= 2 sin (π/4 – γ/2) (cos(α - β )/2 - cos (π/2+ γ)/2)
= 2 sin (π/4 – γ/2) (2 sin (α - β + π/2+ γ)/4 sin (-α + β +π/2-+γ)/4 (using 12)
= 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) (using 6 and 8)

sin α + sin β + sin γ – 1 = 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) …(B)

from A and B we get
RHS =
(sin α + sin β + sin γ – 1)/ (cos α + cos β + cos γ)
= (4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2))/ (4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2))
= tan (π/4 – γ/2) tan (π/4 – β/2) tan (π/2- α/2)
= tan (π/4- α/2) tan (π/4 – β/2) tan (π/4 – γ/2)
= [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan β/2)(1 + tan γ/2)]/[( (using 13 and rearranging the terms)

= LHS