If $ A $ and $ 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 $.

Academic Mathematics NCERT Class 10

Given:

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

To do:

We have to find \( A+B \).

Solution:

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

This implies,

\( \tan (A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})} \)

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

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

\( =\frac{5}{5} \)

\( =1 \)

\( \Rightarrow \tan (A+B)=\tan 45^{\circ} \) (Since $\tan 45^o=1$)

Therefore, the value of $A+B$ is $45^{\circ}$.

Simply Easy Learning

Updated on: 10-Oct-2022

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