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2014-12-17T22:29:02+05:30

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Let  Sin⁻¹ x = A,    Sin⁻¹ y = B

Sin A = x     Sin B = y
Cos A = √(1-x²)          Cos B = √(1-y²)

    Cos⁻¹ x = C
          =>   Cos C = x  = Sin A = Cos (90-A)
          =>   C = 90 - A     or   90 + A  

  Cos⁻¹ y = D
      => Cos D = y = SIn B
     =>  D = 90 + B or   90 - B

Now given equations are,
          A + B = 2π/3  -- equation 1
          C+D = π/2     --- equation 2

1)     C+D= (90 + A) + (90 + B) = π/2
           A+B = -π/2  not possible

2) C+D = (90+A)+(90-B) = π/2
          A-B = -π/2  -- equation 3
       Hence,  from equation 1 and 3,    A = 15 deg.  B = 105 deg

3)  C+D = (90-A) + (90+B)  = π/2
           B-A = -π/2 ,   then  B =15 deg,  A = 105 deg

4)  C+D = 90-A+90-B =  π/2
           A+B = π/2          this is not possible 

x = Sin A = Sin 15 deg and y = Sin 105 deg or Cos 15 deg 
or,       x = Cos 15  and  y = Sin 15


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