Question

U.16 ufa a + B = è, after hilfts (1 + tan a) (1 + tan B) = 2
If a + B = , then prove that (1 + tan 0) (1 + tan B) = 2​

Answer 1

(1+tanA)(1+tanB)=2

∴tan(A+B)=  

1−tanAtanB

tan A + tan B

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Now 1+tanA+tanB+tanAtanB=2

∴tan A + tan B=1−tanAtanB

∴  

1−tanAtanB

tan A + tan B

​  

 

∴ From (1) & (2)

tan(A+B)=1

∴A+B=  

4

π

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