Question

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Answer 1

$\orange{ \boxed{ \boxed{ \begin{array}{c | c} \underline{ \sf \: Procedure}& \underline{ \sf \: Explanation }\\ \\ \rm {tan}^{ - 1}( \sqrt{ \pink{\frac{ 1- cos \: x}{1 + cos \: x} }} ) & \to \to \to \to \to \to \to\bf \to \: given \\ \\ & \because \bf \: 1 - cos2x = 2 {sin}^{2} x\\ & \bf \: so \: \pink{ 1 - cosx} = \green{2 {sin}^{2} \frac{x}{2}} \\ \rm = {tan}^{ - 1} ( \sqrt{ \green{\frac{2 {sin}^{2} \frac{x}{2} }{2 {cos}^{2} \frac{x}{2} } }} )& \bf \: and \:1 + cos2x = 2 {cos}^{2}x \\ & \bf \: so \: \pink{1 + cosx }= \green{ 2 {cos}^{2} \frac{x}{2} }\\ \\ \rm = {tan}^{ - 1} \sqrt{ {tan}^{2} \frac{x}{2} } & \rm \because \: tanx = \frac{sinx}{cosx} \\ \\ \rm = {tan}^{ - 1} .tan \frac{x}{2} & \\ \\ \rm = \frac{x}{2}& \because \bf \: {tan}^{ - 1} tan \: x = x \\ \\ & \bf \: so \: {tan}^{ - 1} tan \frac{x}{2} = \frac{x}{2} \\ \\ = \sf \: answer\end{array}}}}$