let  α & β be two real roots of the equation k+1tan2x−2.λtanx=1−k, where  k≠−1and  λare real numbers. If  tan2α+β=50 then  the value of  λ is

Given that tanα and tanβ are roots of the equationk+1tan2x−2λ tanx+k−1=0⇒sum of roots = tanα+tanβ=2λk+1and, product of roots = tanαtanβ=k−1k+1Now tanα+β=tanα+tanβ1-tanαtanβ=2λk+11−k−1k+1=2λ2=λ2⇒tan2α+β=λ22=50⇒λ=10