Question

plz solve this question guyzz I'll give uhh brain mark ​

Answer 1

$\underline{\underline{\bf{{ prove \: that}}}} : - \\ \\ {tan}^{ - 1} x + {tan}^{ - 1} \frac{2x}{1 - {x}^{2} } = {tan}^{ - 1} \frac{3x - {x}^{3} }{1 - 3 {x}^{2} } \\ \\ \underline{ \underline{\bf{step - by - step \: explanation}}} : -$

Formula used→

$\implies \: {tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} \frac{x + y}{1 - xy}$

Now , taking left hand side ( LHS)

$\rightarrow \: {tan}^{ - 1} x + {tan}^{ - 1} \frac{2x}{1 - {x}^{2} } \\ \\ applying \: the \: given \: formula \: \\ \\ \rightarrow \: {tan}^{ - 1} \bigg( \frac{x + \frac{2x}{1 - {x}^{2} } }{1 - x \times \frac{2x}{1 - {x}^{2} } } \bigg) \\ \\ \rightarrow \: {tan}^{ - 1} \bigg( \frac{ \frac{x - {x}^{3} + 2x }{1 - {x}^{2} } }{1 - \frac{2 {x}^{2} }{1 - {x}^{2} } } \bigg) \\ \\ \rightarrow \: {tan}^{ - 1} \bigg( \frac{ \frac{3x - {x}^{3} }{1 - {x}^{2} } }{ \frac{1 - {x}^{2} - 2 {x}^{2} }{1 - {x}^{2} } } \bigg) \\ \\ \rightarrow \: {tan}^{ - 1} \bigg( \frac{3x - {x}^{3} }{1 - 3 {x}^{2} } \bigg)$

LHS = RHS

Hence proved.