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# If A and B are acute angles such that tan A=12,

Question:

If $\mathrm{A}$ and $\mathrm{B}$ are acute angles such that $\tan A=\frac{1}{2}, \tan B=\frac{1}{3}$ and $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$, find $A+B$.

Solution:

Given:

$\tan A=\frac{1}{2}$...(1)

$\tan B=\frac{1}{3}$....(2)

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \ldots \ldots$(3)

Now by substituting the value of $\tan A$ and $\tan B$ from equation (1) and (2) in equation (3)

We get,

$\tan (A+B)=\frac{\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)}{1-\left(\frac{1}{2}\right) \times\left(\frac{1}{3}\right)}$

$=\frac{\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)}{1-\left(\frac{1}{2 \times 3}\right)}$

$=\frac{\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)}{1-\left(\frac{1}{6}\right)}$

$=\frac{\frac{5}{6}}{\frac{5}{6}}$

$=\frac{5}{6} \times \frac{6}{5}$

$=1$

Therefore,

$\tan (A+B)=1$...(3)

Now we know that

$\tan 45^{\circ}=1 \ldots \ldots(4)$

Now by comparing equation (3) and (4)

We get,

$A+B=45^{\circ}$