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Question

Aditya Gupta · Answer

call alpha=x and beta=y note that a^2= sin^2x+sin^2y+2sinx.siny and b^2= cos^2x+cos^2y+2cosx.cosy now a^2+b^2= 1+1+2(sinx.siny+cosx.cosy)= 2(1+cos(x – y))....(1) and b^2 – a^2= cos^2x – sin^2x+cos^2y – sin^2y+2(cosx.cosy – sinx.siny) = cos2x+cos2y+2cos(x+y)= 2cos(x+y)cos(x – y)+2cos(x+y) or b^2 – a^2= 2cos(x+y)[1+cos(x – y)].....(2) divide (2) by (1), we get the reqd value of cos(x+y) (b^2 – a^2)/(b^2+a^2)= cos(x+y) option B KINDLY APPROVE :))

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Please help me to solve it............................................

Please help me to solve it............................................

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