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if sin x + cos Y = A and cos x + sin y = b then tan of x-y/2 is equal to?

if sin x + cos Y = A and cos x + sin y = b then tan of x-y/2 is equal to?
 

Grade:11

1 Answers

Arun
25750 Points
5 years ago
Dear Student, 
tan(a/2) = sin(a/2)/cos(a/2) = 2sin(a/2)cos(a/2)/[2cos^2(a/2)] = sin(a)/(1+cos(a)) 
tan[(x-y)/2] = sin(x-y)/[1+cos(x-y)] ......(1) 

From (sinx+cosy)^2 + (cosx+siny)^2 = a^2 + b^2, we can get 
2 + 2sinx cosy + 2cosx siny = a^2 + b^2 
2[1+sin(x+y)] = (a^2+b^2) ......(2) 

From (sinx+cosy)^2 - (cosx+siny)^2 = a^2 - b^2, we can get 
2sin(x-y) - cos2x + cos2y = a^2 - b^2 
2sin(x-y) + 2sin(x+y)sin(x-y) = a^2 - b^2 
2[1+sin(x+y)] sin(x-y) = a^2 - b^2 ......(3) 

Combine (2) and (3), and solve for sin(x-y), 
sin(x-y) = (a^2 - b^2)/(a^2 +b^2) ......(4) 

cos^2(x-y) = 1 - sin^2(x-y) = 1 - [(a^2 - b^2)/(a^2 +b^2)]^2 = 4a^2b^2/(a^2 + b^2)^2 
So, cos(x-y) = 2ab/(a^2+b^2) ......(5) 

Substitute (4) and (5) into (1), 
tan[(x-y)/2] 
= sin(x-y)/[1+cos(x-y)] 
= [(a^2 - b^2)/(a^2 +b^2)]/[1+2ab/(a^2+b^2)] 
= (a^2-b^2)/(a+b)^2
= (a-b)/(a+b)

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