Extremely challenging trigonometry conditional proof?

Question

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)

Answer 1

I have shown the whole proof in the e-mail i sent u!!

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Answer 2

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. A B are used for generic ones and α  ÃŽÂ²  ÃŽÂ³ for these specified in problems)

(actually they should be derived if required as required but I am deriving to keep the flow)

α + β + γ = π/2

so α + β  = π/2 – γ

or  ÃŽÂ± + 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 -  ÃŽÂ²/2 …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 -  ÃŽÂ²/2) cos (α - β - π/2 - γ)/2 (using 6)

= 4 cos (π/4 – γ/2 ) cos (π/4 -  ÃŽÂ²/2) cos (π/4 – α/2) using 7

So cos α + cos β + cos   γ = 4 cos (π/4 – γ/2 ) cos (π/4 -  ÃŽÂ²/2) 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  Ã¢Â€Â¦ 9

sin (α +  ÃŽÂ² +   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)  Ã¢Â€Â¦ 13

based on above numerator

sin  ÃŽÂ± + sin  ÃŽÂ² + sin  ÃŽÂ³ – 1

= 2 sin (α + β)/2 cos(α - β )/2 + sin  ÃŽÂ³ – sin (α +  ÃŽÂ² +   γ) (using 9 and 10)

= 2 sin (π/4 – γ/2) cos(α - β )/2 –(sin (α +  ÃŽÂ² +   γ) - sin  ÃŽÂ³) (using 5)

= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (α +  ÃŽÂ²)/2  cos (α +  ÃŽÂ² +   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  ÃŽÂ± + sin  ÃŽÂ² + sin  ÃŽÂ³ – 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  ÃŽÂ± + sin  ÃŽÂ² + sin  ÃŽÂ³ – 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 -  ÃŽÂ²/2) 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