INVERSE TRIGONOMETRIC FUNCTIONS (Revision)

\[1.\ Show\ that\ ( sin α + sin β )^2\ +\ ( cos α – cos β )^2\ =\ 4\ sin^2 (\frac{α\ +\ β}{2})\ \hspace{15cm}\]

\[\color {black}{Solution:}\ \hspace{19cm}\]

\[W.\ K.\ T\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ and\]

\[Cos\ A\ -\ Cos\ B =\ -\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\]

\[L.\ H.\ S\ =\ ( sin α + sin β )^2\ +\ ( cos α – cos β )^2\ \hspace{10cm}\]

\[=\ [2\ Sin\ (\frac{α\ +\ β}{2})\ Cos\ (\frac{α\ -\ β}{2})]^2\ +\ [-\ 2\ Sin\ (\frac{α\ +\ β}{2})\ Sin\ (\frac{α\ -\ β}{2})]^2\ \hspace{10cm}\]

\[=\ 4\ Sin^2\ (\frac{α\ +\ β}{2})\ Cos^2\ (\frac{α\ -\ β}{2})\ +\ 4\ Sin^2\ (\frac{α\ +\ β}{2})\ Sin^2\ (\frac{α\ -\ β}{2})\ \hspace{10cm}\]

\[=\ 4\ Sin^2\ (\frac{α\ +\ β}{2})[Cos^2\ (\frac{α\ -\ β}{2})\ +\ Sin^2\ (\frac{α\ -\ β}{2})]\ \hspace{10cm}\]

\[=\ 4\ Sin^2\ (\frac{α\ +\ β}{2})\ (1)\ = 4\ Sin^2\ (\frac{α\ +\ β}{2})\ =\ R.H.S\ \hspace{10cm}\]

\[2.\ Show\ that\ 2\ Tan^{-1}\ (\frac{2}{3})\ =\ Tan^{-1}\ (\frac{12}{5})\ \hspace{15cm}\]

\[\color {black}{Solution:}\ \hspace{19cm}\]

\[L.\ H.\ S\ =\ 2\ Tan^{-1}\ (\frac{2}{3})\ \hspace{10cm}\]

\[=\ Tan^{-1}\ (\frac{2}{3})\ +\ Tan^{-1}\ (\frac{2}{3})\ \hspace{10cm}\]

\[=\ Tan^{-1}\ (\frac{\frac{2}{3}\ +\ \frac{2}{3}}{1\ -\ \frac{2}{3}\ ×\ \frac{2}{3}})\ \hspace{2cm}\ \because\ Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})\]

\[=\ Tan^{-1}\ (\frac{\frac{2\ +\ 2}{3}}{1\ -\ \frac{4}{9}})\ \hspace{10cm}\]

\[=\ Tan^{-1}\ (\frac{\frac{4}{3}}{\frac{5}{9}})\ \hspace{10cm}\]

\[=\ Tan^{-1}\ (\frac{4}{3}\ ×\ \frac{9}{5})\ \hspace{10cm}\]

\[=\ Tan^{-1}\ (\frac{12}{5})\ =\ R.H.S\ \hspace{10cm}\]