Question

plzz solve this equation

Answer 1

➡HERE IS YOUR ANSWER⬇

$\frac{1 - cosx}{1 + cosx} \\ \\ = \frac{1 - (1 - 2 {sin}^{2} \frac{x}{2}) }{1 + (2 {cos}^{2} \frac{x}{2} - 1) } \\ \\ = \frac{2 {sin}^{2} \frac{x}{2} }{2 {cos}^{2} \frac{x}{2} } \\ \\ = \frac{ ({sin \frac{x}{2} })^{2} }{( {cos \frac{x}{2} })^{2} } \\ \\ = ( {tan \frac{x}{2}) }^{2}$

So,

${tan}^{ - 1} \sqrt{ \frac{1 - cosx}{1 + cosx} } \\ \\ = {tan}^{ - 1} \sqrt{( {tan} \frac{x}{2} )^{2}} \\ \\ = {tan}^{ - 1} (tan \frac{x}{2} ) \\ \\ = \frac{x}{2}$

Rule used :

${tan}^{ - 1} (tanx) = x$

⬆HOPE THIS HELPS YOU⬅