If `tanalpha=1/7,sinbeta=1/(sqrt(10)),` prove that`alpha+2beta=pi/4` , where `0

Explanation

`tan (alpha+2beta)=(tan alpha+tan 2 beta)/(1-tan alpha tan 2 beta)=((1)/(7)+tan 2 beta)/(1-(1)/(7)tan2 beta)`
Now, `tan 2 beta=(2 tan beta)/(1-tan^(2)beta)=(2xx(1)/(3))/(1-(1)/(9))=(3)/(4)`
`[tan betagt0 "as" 0ltbetaltpi//2]`
Substituting the value of `tan 2 beta` in Eq. (i). we get
`tan (alpha+2beta)=((1)/(7)+(3)/(4))/(1-(1)/(7)xx(3)/(4))=(25)/(25)=1`
Now, `0ltalphalt(pi)/(2)"and"0ltbetalt(pi)/(2)`
`therefore 0lt2betaltpi`, but `tan 2 beta=(3)/(4)gt0`.
`rArr 0lt 2betalt(pi)/(2)`
Hence `0ltalpha+2betaltpi`.
In the interval `(0,pi) tan theta` takes value 1 at `pi//4`. only. Therefore,
`alpha+2beta=(pi)/(4)`.