If $(1 + tan A) . (1 + tan B) = 2$ , then $A + B$ is

\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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\frac{π}{6}

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Solution

The correct option is C $\frac{π}{4}$
$(1 + tan A) (1 + tan B) = 2$
$∴ tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ ........... $(1)$
Now $1 + tan A + tan B + tan A tan B = 2$
$∴ tan A + tan B = 1 - tan A tan B$
$∴ \frac{tan A + tan B}{1 - tan A tan B} = 1$ ........ $(2)$
$∴$ From $(1)$ & $(2)$
$tan (A + B) = 1$
$∴ A + B = \frac{π}{4}$

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If (1+tanA).(1+tanB)=2, then A+B is

The correct option is C π4(1+tanA)(1+tanB)=2∴tan(A+B)=tanA+tanB1−tanAtanB ........... (1)Now 1+tanA+tanB+tanAtanB=2∴tanA+tanB=1−tanAtanB∴tanA+tanB1−tanAtanB=1 ........ (2)∴ From (1) & (2)tan(A+B)=1∴A+B=π4

If $(1 + tan A) . (1 + tan B) = 2$ , then $A + B$ is