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B
π3
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C
π2
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D
π
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Solution
The correct option is Dπ4 Given, tanα=aa+1 and tanβ=12a+1 ∴tan(α+β)=tanα+tanβ1−tanαtanβ =aa+1+12a+11−aa+1×12a+1 =a(2a+1)+(a+1)(a+1)(2a+1)−a =2a2+2a+12a2+2a+1=1⇒α+β