# INVERSE TRIGONOMETRIC FUNCTIONS (Excercise)

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {1.}\color {red} {Prove\ that}\ Sin\ 50^{0}\ -\ Sin\ 70^{0}\ +\ Sin\ 10^{0}\ =\ 0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T.\ Sin\ A\ -\ Sin\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$L.\ H.\ S\ =\ Sin\ 50^{0}\ -\ Sin\ 70^{0}\ +\ Sin\ 10^{0}\ \hspace{10cm}$
$=\ 2\ Cos\ (\frac{50^{0}\ +\ 70^{0}}{2})\ Sin\ (\frac{50^{0}\ -\ 70^{0}}{2})\ +\ Sin\ 10^{0}\ \hspace{10cm}$
$=\ 2\ Cos\ 60^{0}\ Sin\ (-10^{0})\ +\ Sin\ 10^{0}\ \hspace{10cm}$
$=\ 2\ ×\ \frac{1}{2}\ (-\ Sin\ 10^{0}\ +\ Sin\ 10^{0})\ \hspace{10cm}$
$=\ 0\ =\ R.H.S\ \hspace{10cm}$
$\color {purple} {2}.\ \color {red} {Prove\ that}\ Cos\ 80^{0}\ +\ Cos\ 40^{0}\ -\ Cos\ 20^{0}\ =\ 0\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T.\ Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ \hspace{18cm}$
$L.\ H.\ S\ =\ Cos\ 80^{0}\ +\ Cos\ 40^{0}\ -\ Cos\ 20^{0}\ \hspace{10cm}$
$=\ 2\ Cos\ (\frac{80^{0}\ +\ 40^{0}}{2})\ Cos\ (\frac{80^{0}\ -\ 40^{0}}{2})\ -\ Cos\ 20^{0}\ \hspace{10cm}$
$=\ 2\ Cos\ 60^{0}\ Cos\ (-20^{0})\ -\ Cos\ 20^{0}\ \hspace{10cm}$
$=\ 2\ ×\ \frac{1}{2}\ (Cos\ 20^{0}\ -\ Cos\ 20^{0})\ \hspace{10cm}$
$=\ 0\ =\ R.H.S\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {3}.\ \color {red} {Prove\ that}\ Sin\ 20^{0}\ Sin\ 40^{0}\ Sin\ 60^{0}\ Sin\ 80^{0}\ =\ \frac{3}{16}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Sin\ (A\ +\ B)Sin\ (A\ -\ B)\ =\ Sin^2\ A\ -\ Sin^2\ B\ and$
$Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A$
$L.\ H.\ S\ =\ Sin\ 20^{0}\ Sin\ 40^{0}\ Sin\ 60^{0}\ Sin\ 80^{0}\ \hspace{10cm}$
$L.\ H.\ S\ =\ \frac{\sqrt{3}}{2}\ Sin\ 20^{0}\ Sin\ 40^{0}\ Sin\ 80^{0}\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{2}\ Sin\ 20^{0}\ Sin\ (60^{0}\ -\ 20^{0}) Sin\ ((60^{0}\ +\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{2}\ Sin\ 20^{0}\ (Sin^2\ 60^{0}\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{2}\ Sin\ 20^{0}\ ((\frac{\sqrt{3}}{2})^2\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{2}\ Sin\ 20^{0}\ (\frac{3}{4}\ -\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ Sin\ 20^{0}\ (3\ -\ 4\ Sin^2\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}(3\ Sin\ 20^{0}\ -\ 4\ Sin^3\ 20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ Sin\ 3(20^{0})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ Sin\ 60^{0}\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ (\frac{\sqrt{3}}{2})\ \hspace{10cm}$
$=\ \frac{3}{16}\ =\ R.H.S\ \hspace{10cm}$
$\color {purple} {4.}\ \color {red} {Prove:}\ Tan^{-1}\ (\frac{3x\ -\ x^3}{1\ -\ 3\ x^2})\ =\ 3\ Tan^{-1}\ x\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{19cm}$
$Put\ x\ =\ Tan\ θ,\ \implies\ θ\ =\ Tan^{-1}\ x\ \hspace{10cm}$
$L.\ H.\ S\ =\ Tan^{-1}\ (\frac{3x\ -\ x^3}{1\ -\ 3\ x^2})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{3\ Tan\ θ\ -\ Tan^3\ θ}{1\ -\ 3\ Tan^2\ θ})\ \hspace{10cm}$
$=\ Tan^{-1}\ (Tan\ 3θ)\ \hspace{10cm}$
$=\ 3\ θ\ \hspace{10cm}$
$=\ 3\ Tan^{-1}\ x\ =\ R.H.S\ \hspace{10cm}$
$\color {purple} {5.}\ \color {red} {Show\ that}\ Tan^{-1}\ (\frac{4}{3})\ – Tan^{-1}\ (\frac{1}{7})\ =\ \frac{\pi}{4}\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{19cm}$
$L.\ H.\ S\ =\ Tan^{-1}\ (\frac{4}{3})\ – Tan^{-1}\ (\frac{1}{7})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{\frac{4}{3}\ -\ \frac{1}{7}}{1\ +\ \frac{4}{3}\ ×\ \frac{1}{7}})\ \hspace{2cm}\ \because\ Tan^{-1}\ x\ – Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ -\ y}{1\ +\ x\ y})$
$=\ Tan^{-1}\ (\frac{\frac{28\ -\ 3}{21}}{\frac{21\ +\ 4}{21}})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{\frac{25}{21}}{\frac{25}{21}})\ \hspace{10cm}$
$=\ Tan^{-1}\ (1)\ =\ \frac{\pi}{4}\ =\ R.H.S\ \hspace{10cm}$
$\color {purple} {6.}\ \color {red} {Show\ that}\ 2\ Tan^{-1}\ (\frac{2}{3})\ =\ Tan^{-1}\ (\frac{12}{5})\ \hspace{15cm}$
$\color {blue}{Solution:}\ \hspace{19cm}$
$L.\ H.\ S\ =\ 2\ Tan^{-1}\ (\frac{2}{3})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{2}{3})\ +\ Tan^{-1}\ (\frac{2}{3})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{\frac{2}{3}\ +\ \frac{2}{3}}{1\ -\ \frac{2}{3}\ ×\ \frac{2}{3}})\ \hspace{2cm}\ \because\ Tan^{-1}\ x\ + Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ +\ y}{1\ -\ x\ y})$
$=\ Tan^{-1}\ (\frac{\frac{2\ +\ 2}{3}}{\frac{9\ -\ 4}{9}})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{\frac{4}{3}}{\frac{5}{9}})\ \hspace{10cm}$
$=\ Tan^{-1}\ (\frac{12}{5})\ =\ R.H.S\ \hspace{10cm}$