Question

if tanA= 5/6 and tanB=1/11 then find the value of (A+B)

Answer 1

$<b>$
$tan(A+B) = \dfrac{tanA+tanB}{1-tanA \: tanB} \: \\ \\ = \frac{ \dfrac{5}{6} + \dfrac{1}{11} }{ 1 - \dfrac{5}{6} \times \dfrac{1}{11} } \\ \\ = \frac{ \dfrac{55 + 6}{66} }{1 - \dfrac{5}{66} } \\ \\ = \frac{ \dfrac{61}{66} }{ \dfrac{ 66 - 5}{66} } \\ \\ = \dfrac{ \dfrac{61}{66} }{ \dfrac{61}{65} } \\ \\ = \dfrac{61}{65} \times \dfrac{65}{61} \\ \\ = 1$
tan(A+B) = 1 = tan 45°

$\bf{A+B = 45}$

A+B = π/4

$\bf{Thanks}$

Have a colossal day ahead

Be Brainly

Answer 2

Answer:

The value of tan (A+B) is 1.

Step-by-step explanation:

Given

tanA=5/6

tanB=1/11

To find the value of tan(A+B)

Solution:

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

=$\frac{5}{6} +\frac{1}{11} /1-\frac{5}{6}.1/11$

=55+6/66 / 1-5/66

=61/66 / 66-5/66    

=$\frac{61}{66} x\frac{66}{61}$

=1

Thus, the value of tan(A+B) is 1.