If `sin(alpha+beta)=1` and `sin(alpha-beta)=1/2`, where `0lt=alpha, betalt= pi/2` , then find the values of `tan(alpha+2beta)` and `tan(2alpha+beta)` .

Explanation

Given,`sin(alpha+beta)=1,sin(alpha−beta)=1/2`

or,`alpha+beta=pi/2,alpha−beta=π/6`

solving we get


`=>alpha=pi/3​,beta=pi/6​`

`tan(alpha+2beta)`

=`tan(pi/3+pi/3)`


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