Question

Prove that.
$tan^{-1}[ \sqrt{\frac{1-cosx}{1+cosx}} ] = \frac{x}{2}$

Answer 1

Formulas used:

$1 - cos \: x = 2 \: {sin}^{2} ( \frac{x}{2} ) \\ \\ \\ 1 + cos \: x = 2 \: {cos}^{2} ( \frac{x}{2} )$
we have

LHS =
${tan}^{ - 1} ( \sqrt{ \frac{1 - cos \: x}{1 + cos \: x} } ) \\ \\ \\ = {tan}^{ - 1} \sqrt{ \frac{2 {sin}^{2} ( \frac{x}{2}) }{2 \: {cos}^{2}( \frac{x}{2} ) } } \\ \\ \\ = {tan}^{ - 1} \sqrt{ {tan}^{2}( \frac{x}{2} ) } \\ \\ \\ = {tan}^{ - 1} (tan \: \frac{x}{2} ) \\ \\ = \frac{x}{2}$
= RHS

HENCE PROVED

Answer 2

Formulas used:

$\begin{gathered} 1 - cos \:x= 2 \: {sin}^{2} ( \frac{x}{2} ) \\ \\\\ 1 + cos \: x = 2 \: {cos}^{2} ( \frac{x}{2} ) \\ \\\end{gathered}$

we have

$LHS =\begin{gathered} {tan}^{ - 1} ( \sqrt{ \frac{1 - cos \: x}{1 + cos \: x} } ) \\ \\ \\ ={tan}^{ - 1} \sqrt{ \frac{2 {sin}^{2} ( \frac{x}{2}) }{2 \: {cos}^{2}( \frac{x}{2} ) } } \\ \\ \\ = {tan}^{ - 1} \sqrt{ {tan}^{2}( \frac{x}{2} ) } \\ \\ \\ = {tan}^{ - 1} (tan \: \frac{x}{2} ) \\ \\ = \frac{x}{2} \end{gathered}$

HENCE PROVED