Question

In a triangle ABC if (a^2+b^2)/(a^2-b^2) sin(A-B)=1 and C is not right angle then cos(A-B)=

1) tan(C/2 + pi/4)
2) tan(C/2 - pi/4)
3) cos(C/2 + pi/4)
4) sin(C/2 - pi/4)

Answer 1

Given :   In a triangle ABC if (a^2+b^2)/(a^2-b^2) sin(A-B)=1 and C is not right angle  

To Find :   cos(A-B)

1) tan(C/2 + pi/4)

2) tan(C/2 - pi/4)

3) cos(C/2 + pi/4)

4) sin(C/2 - pi/4)

Solution:

(a² + b²) Sin (A - B) / (a² - b²)  = 1

a/SinA = b/sinB = c/SinC  = k

=> a = kSinA ,  b = k SinB

=> (k²sin²A + k²Sin²B) Sin (A - B) /(k²sin²A - k²Sin²B)  = 1

=> ( sin²A +  Sin²B) Sin (A - B) /  (sin²A -  Sin²B)  = 1

=>  ( 2sin²A +  2Sin²B) Sin (A - B) /  (2sin²A -  2Sin²B)  = 1

=> ( 1 - Cos2A + 1 - Cos2B) Sin (A - B) / ( 1 - Cos2A - (1 - Cos2B)) = 1

=>   ( 2 - (Cos2A + Cos2B)) Sin (A - B)/ ( Cos2B - Cos2A)  = 1

=> ( 2 -  2Cos(A + B)Cos(A - B) Sin(A-B) / (2Sin(A + B)Sin (A - B) ) = 1

=>  ( 1 -   Cos(A + B)Cos(A - B) )   /   Sin(A + B)  = 1

=> {1 - Cos (π - C) Cos(A - B)) / Sin (π - C)  = 1

=> 1   - Cos (π - C) Cos(A - B) =  Sin (π - C)

=>  Cos(A - B) =  ( 1 - Sin (π - C) ) / Cos (π - C)

=>  Cos(A - B) =  ( 1 - Sin (C) ) / -Cos (C)

=> Cos(A - B) = - ( cos²C/2 + Sin²C/2 - 2SinC/2.CosC/2 ) / (Cos² C/2 - Sin²C/2)

=> Cos(A - B) = - ( cos C/2 - Sin C/2)² / (Cos² C/2 - Sin²C/2)

=> Cos(A - B) = - ( cos C/2 - Sin C/2)  / (Cos  C/2 + Sin C/2)

=> Cos(A - B) = - ( 1 - tan C/2)  / (1   + tan C/2)

=> Cos(A - B) = -  tan (π/4 - C/2)

=> Cos(A - B) =   tan (  C/2 -π/4 )

Hence option 2 is correct

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