\[\color {purple} {Example\ 1:}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}\]
\[\frac{Tan\ {20}^0\ +\ Tan\ {25}^0}{1\ -\ Tan\ {20}^0\ Tan\ {25}^0}\ =\ Tan({20}^0\ +\ {25}^0)\ =\ Tan{45}^0\ =\ 1\ \hspace{10cm}\]
\[\color {purple} {Example\ 5:}\ If\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21},\ \color {red} {Show\ that\ A\ +\ B\ =\ {45}^0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21}\ \hspace{18cm}\]
\[W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}\]
\[=\ \frac{\frac{10}{11}\ +\ \frac{1}{21}}{1\ -\ \frac{10}{11}\ ×\ \frac{1}{21}}\ \hspace{10cm}\]
\[=\ \frac{\frac{210\ +\ 11}{231}}{1\ -\ \frac{10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{231\ -\ 10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{221}{231}}\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ {45}^0\ \hspace{10cm}\]
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