COMPOUND ANGLES (REVISION)

\[1.\ If\ Tan\ A\ =\ \frac{5}{6} \ and\ Tan\ B\ =\ \frac{1}{11},\ Show\ that\ A\ +\ B\ =\ \frac{π}{4}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Tan\ A\ =\ \frac{5}{6} \ and\ Tan\ B\ =\ \frac{1}{11}\ \hspace{18cm}\]
\[W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}\]
\[=\ \frac{\frac{5}{6}\ +\ \frac{1}{11}}{1\ -\ \frac{5}{6}\ ×\ \frac{1}{11}}\ \hspace{10cm}\]
\[=\ \frac{\frac{55\ +\ 6}{66}}{1\ -\ \frac{5}{66}}\ \hspace{10cm}\]
\[=\ \frac{\frac{61}{66}}{\frac{66\ -\ 5}{66}}\ \hspace{10cm}\]
\[=\ \frac{\frac{61}{66}}{\frac{61}{66}}\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ \frac{π}{4}\ \hspace{10cm}\]
\[2.\ If\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14},\ Prove\ that\ A\ -\ B\ =\ \frac{π}{3}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Given\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A\ -\ B)\ =\ Cos A\ Cos B\ +\ Sin A\ Sin B\ \hspace{15cm}\]
\[Sin\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Sin\ A\ =\ \sqrt{1\ -\ Cos^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{7})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{49\ -\ 1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{48}{49}}\ \hspace{10cm}\]
\[=\ \frac{\sqrt{48}}{7}\ \hspace{10cm}\]
\[Sin\ A\ =\ \frac{4\ \sqrt{3}}{7}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{13}{14})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{196\ -\ 169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{27}{196}}\ \hspace{10cm}\]
\[=\ \frac{\sqrt{27}}{14}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{3\ \sqrt{3}}{14}\ \hspace{10cm}\]
\[Cos( A\ -\ B)\ =\ (\frac{1}{7})\ (\frac{13}{14})\ +\ (\frac{4\ \sqrt{3}}{7}\ (\frac{3\ \sqrt{3}}{14})\ \hspace{10cm}\]
\[=\ \frac{13}{98}\ +\ \frac{12\ (3)}{98}\ \hspace{10cm}\]
\[=\ \frac{13}{98}\ +\ \frac{36}{98}\ \hspace{10cm}\]
\[=\ \frac{13\ +\ 36}{98}\ \hspace{10cm}\]
\[=\ \frac{49}{98}\ \hspace{10cm}\]
\[Cos( A\ -\ B)\ =\ \frac{1}{2}\ \hspace{10cm}\]
\[(A\ -\ B)\ =\ \frac{π}{3}\ \hspace{10cm}\]