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If sin x = 3/5 , cos y = − 12/13 , where x and y both lie in second quadrant, find the value of sin (x + y).

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We know that  
sin (x + y) = sin x cos y + cos x sin y ... (1)  
Now cos2x = 1 – sin2x = 1 – 9/25 = 16/25  
Therefore cos x = ± 4/5.  
Since x lies in second quadrant, cos x is negative.  
Hence cos x = −4/5  
Now sin2y = 1 – cos2y = 1 – 144/169 = 25/169  
i.e. sin y = ± 5/13.  
Since y lies in second quadrant, hence sin y is positive

Therefore, sin y =5/13. 

Substituting the values of sin x, sin y, cos x and cos y in (1), we get  
sin(x + y) 3/5 × (-12/13) + (−4/5) × 5/13 = (-36/65) –(20/65) = -56/65

Therefore, sin(x+y) = -56/65

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