Question

If A + B = 45° , find (1 + tan A)
(1 + tab B).​

Answer 1

Answer:

A+B=45

tan(A+B)= tan45

tanA+tanB/1- tanAtanB=1

 tanA+ tanB= 1-tanAtanB 

tanA +tanB+ tanAtanB=1

 1+tanA +tanB+ tanAtanB=1+1  

        (adding 1 on both sides)

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

 (1+tanA)(1+tanB)=2

hence proved

Answer 2

Answer:

$\huge\bf\underline\blue{AnSweR:}$

$\tan(a + b) = \tan45 = 1$

$\frac{ \tan \: a + \tan \: b }{1 - \tan \: a \: \tan \: b } = 1$

$\tan \: a + \tan \: b + \tan \: a \tan \: b = 1$

$1 + \tan \: a + \tan \: b + \tan \: a \tan \: b = 1 + 1$

$\implies\bf\green{(1 + tan A)(1+ tan B) = 2}$