Question

If tan (A+B) = √3 and tanA=1 ,then tan∆B=

Answer 1

Answer:

Step-by-step explanation:

$tan(A+B)=\sqrt{3} \\\\\\tan(A+B)=\dfrac{tanA+tanB}{1-tanAtanB} \\\\\\\\\dfrac{tanA+tanB}{1-tanAtanB}=\sqrt{3} \\\\\\\\\dfrac{1+tanB}{1-1*tanB}=\sqrt{3} \\\\\\\\1+tanB}=\sqrt{3} (1-tanB)\\\\\\1+tanB=\sqrt{3} -\sqrt{3} tanB\\\\\\tanB+\sqrt{3} tanB=\sqrt{3} +1\\\\\\tanB(\sqrt{3} +1)=(\sqrt{3} +1)\\\\\\tanB=\dfrac{\sqrt{3} +1}{\sqrt{3} +1 } \\\\\\\\tanB=1$

Answer 2

GIVEN:

tan(A+B)=√3

tanA=1

TO FIND :

tan B

FORMULA USED :

$tan(A+B)= \frac{tanA + tanB}{1 - tanAtanB}$

SOLUTION :

tan(A+B)=√3

tanA=1

Now by using the formula =>

$⟶tan(A+B)= \frac{tanA + tanB}{1 - tanAtanB}$

put the given values:-

$⟶ \sqrt{3} = \frac{1+ tanB}{1 -( 1 \times tanB)}$

now cross multiply :-

$⟶ \sqrt{3} (1 - tanB )= 1+ tanB$

$⟶ \sqrt{3} - \sqrt{3}tanB = 1+ tanB$

$⟶ \sqrt{3} - 1 = tanB + \sqrt{3}tanB$

$⟶ \sqrt{3} - 1 = (1 + \sqrt{3})tanB$

$⟶ \frac{ \sqrt{3} - 1}{ (1 + \sqrt{3})} =tanB$

ANSWER :

$⟶ \frac{ \sqrt{3} - 1}{ (1 + \sqrt{3})} =tanB$

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Other Formulae :-

1. Sin(A+B) = Sin A CosB + Cos A SinB

2. Cos (A+B) = Cos A CosB - Sin ASinB