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Question

If 0<x,y<π and cosx+cosycos(x+y)=32, then sinx+cosy is equal to :

A
1+32
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B
132
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C
32
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D
12
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Solution

The correct option is A 1+32
cosx+cosycos(x+y)=32
2cos(x+y2)cos(xy2)[2cos2(x+y2)1]=32
2cos(x+y2)cos(xy2)2cos2(x+y2)=12
4cos(x+y2)cos(xy2)4cos2(x+y2)=1=cos2(xy)2+sin2(xy2)
4cos2(x+y2)cos2(xy2)4cos(x+y2)cos(xy2)+sin2(xy2)=0
(cos(xy2)2cos(x+y2))2+sin2(xy2)=0
sinxy2=0x=y
and cosxy2=2cosx+y2
cosx=12=cosy
sinx+cosy=32+12=3+12

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