α and β are the positive acute angles and satisfying equations 5sin2β=3sin2α and tanβ=3tanα simultaneously. Then the value of tanα+tanβ is
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Solution
Given tanβ=3tanα Also given 5sin2β=3sin2α ⇒5(2tanβ1+tan2β)=3(2tanα1+tan2α) ⇒30tanα1+9tan2α=6tanα1+tan2α ⇒24tanα−24tan3α=0 ⇒tanα=−1,0,1 Since,α,β are positive acute angles tanα=1 ⇒tanβ=3 ⇒tanα+tanβ=4