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If alphaa n dbeta are acute angles such ...

If `alphaa n dbeta` are acute angles such that `t a nalpha=m/(m+1)a n dt a nbeta=1/(2m+1)` , prove that `alpha+beta=pi/4` .

Text Solution

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`tan(A+B)=(tanA+tanB)/(1-tanAtanB)`
`=(m/(m+1)+1/(2m+1))/(1-m/(m+1).1/(2m+1))`
`=((m(2m+1)+(m+1))/((m+1)(2m+1)))/(((m+1)(2m+1)-m)/((m+1)(2m+1)))`
`(2m^(2)+m+m+1)/(2m^(2)+2m+m+1-m)`
`=(2m^(2)+2m+1)/(2m^(2)+2m+1)=1=tanpi/4`
`therefore A+B=pi/4`. Hence Proved.
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