If α+β=π2 and β+γ=α, then tanα equals [2001]
(3)
Given: α+β=π2⇒α=π2-β
⇒tanα=tan(π2-β)=cotβ=1tanβ
⇒tanαtanβ=1⇒1+tanαtanβ=2
Now, tan(α-β)=tanα-tanβ1+tanαtanβ
⇒tanγ=tanα-tanβ2
⇒2tanγ=tanα-tanβ⇒tanα=2tanγ+tanβ