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Grade: 7

                        

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

3 months ago

Answers : (1)

Aditya Gupta
2065 Points
							
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 :))
3 months ago
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