Question

if tan(A+B)=√3 and tanA=1 then find angle B​

Answer 1

Step-by-step explanation:

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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 :-

$⟶ \sqrt{3} (1 - tanB )= 1+ tanB$

$⟶ \sqrt{3} - \sqrt{3}tanB = 1+ tanB$

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

$⟶ \sqrt{3} - 1 = (1 + \sqrt{3})tanB$

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

we know :-

$⟶ \frac{ \sqrt{3} - 1}{ (1 + \sqrt{3})} =tan15°$

Therefore , TanB = tan15°

therefore angle B = 15°

ANSWER :

angle B = 15°

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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

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