Question

$\begin{gathered} \rm2 {tan}^{ - 1} \bigg \{tan \frac{ \alpha }{2} \: . \: tan \bigg( \frac{\pi}{4} - \frac{ \beta }{2} \bigg) \bigg \} = {tan}^{ - 1} \bigg \{ \frac{sin \alpha \: . \: cos \beta }{cos \alpha + sin \beta } \bigg \} \\ \\ \end{gathered}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm2 {tan}^{ - 1} \bigg \{tan \frac{ \alpha }{2} \: . \: tan \bigg( \frac{\pi}{4} - \frac{ \beta }{2} \bigg) \bigg \}$

We know,

$\boxed{ \rm{ \: {2tan}^{ - 1}x = {tan}^{ - 1}\bigg(\dfrac{2x}{1 - {x}^{2} } \bigg) \: }}$

So, using this result, we get

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}tan\bigg(\dfrac{\pi}{4} - \dfrac{ \beta }{2} \bigg) }{1 - {tan}^{2} \dfrac{ \alpha }{2} {tan}^{2}\bigg(\dfrac{\pi}{4} - \dfrac{ \beta }{2}\bigg) }\bigg \}$

We know,

$\boxed{ \rm{ \:tan\bigg(\dfrac{\pi}{4} - x\bigg) = \frac{1 - tanx}{1 + tanx} \: }}$

So, using this identity, we get

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(\dfrac{1 - tan\dfrac{ \beta }{2}}{1 + tan\dfrac{ \beta }{2}} \bigg) }{1 - {tan}^{2} \dfrac{ \alpha }{2} {\bigg[\dfrac{1 - tan\dfrac{ \beta }{2}}{1 + tan\dfrac{ \beta }{2}} \bigg]}^{2} }\bigg \}$

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(1 - tan\dfrac{ \beta }{2}\bigg) \bigg(1 + tan\dfrac{ \beta }{2} \bigg) }{ {\bigg(1 + tan\dfrac{ \beta }{2} \bigg)}^{2} - {tan}^{2}\dfrac{ \alpha }{2} {\bigg(1 - tan\dfrac{ \beta }{2} \bigg)}^{2} } \bigg \}$

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(1 - tan^{2} \dfrac{ \beta }{2}\bigg) }{\bigg(1 + {tan}^{2}\dfrac{ \beta }{2} + 2tan\dfrac{ \beta }{2}\bigg) - {tan}^{2}\dfrac{ \alpha }{2}\bigg(1 + {tan}^{2}\dfrac{ \beta }{2} - 2tan\dfrac{ \beta }{2} \bigg) } \bigg \}$

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(1 - tan^{2} \dfrac{ \beta }{2}\bigg) }{\bigg(1 - {tan}^{2}\dfrac{ \alpha }{2}\bigg) + {tan}^{2}\dfrac{ \beta }{2} \bigg(1 - {tan}^{2}\dfrac{ \alpha }{2} \bigg) +2tan\dfrac{ \beta }{2} \bigg(1 + {tan}^{2} \dfrac{ \alpha }{2} \bigg) } \bigg \}$

$\rm \: = {tan}^{ - 1}\bigg \{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(1 - tan^{2} \dfrac{ \beta }{2}\bigg) }{\bigg(1 - {tan}^{2}\dfrac{ \alpha }{2}\bigg) \bigg(1 + {tan}^{2}\dfrac{ \beta }{2} \bigg) +2tan\dfrac{ \beta }{2} \bigg(1 + {tan}^{2} \dfrac{ \alpha }{2} \bigg) } \bigg \}$

$\rm \: = {tan}^{ - 1} \: \dfrac{\dfrac{2tan\dfrac{ \alpha }{2}\bigg(1 - tan^{2} \dfrac{ \beta }{2}\bigg)}{\bigg(1 + {tan}^{2}\dfrac{ \alpha }{2}\bigg) \bigg(1 + {tan}^{2}\dfrac{ \beta }{2} \bigg) } }{\dfrac{\bigg(1 - {tan}^{2}\dfrac{ \alpha }{2}\bigg) \bigg(1 + {tan}^{2}\dfrac{ \beta }{2} \bigg) +2tan\dfrac{ \beta }{2} \bigg(1 + {tan}^{2} \dfrac{ \alpha }{2} \bigg) }{\bigg(1 + {tan}^{2}\dfrac{ \alpha }{2}\bigg) \bigg(1 + {tan}^{2}\dfrac{ \beta }{2} \bigg) } }$

can be rewritten as

$\rm \: = {tan}^{ - 1} \: \dfrac{\dfrac{2tan\dfrac{ \alpha }{2}}{1 + {tan}^{2} \dfrac{ \alpha }{2}} \times \dfrac{1 - {tan}^{2} \dfrac{ \beta }{2}}{1 + {tan}^{2} \dfrac{ \beta }{2}} }{\dfrac{1 - {tan}^{2} \dfrac{ \alpha }{2}}{1 + {tan}^{2}\dfrac{ \alpha }{2} } + \dfrac{2tan\dfrac{ \beta }{2}}{1 + {tan}^{2} \dfrac{ \beta }{2}} }$

We know,

$\boxed{ \rm{ \:sin2x = \frac{2tanx}{1 + {tan}^{2} x} \: }}$

and

$\boxed{ \rm{ \:cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x} \: }}$

So, using these results, we have

$\rm \: = {tan}^{ - 1} \: \bigg( \dfrac{sin \alpha \: cos \beta }{cos \alpha + sin \beta } \bigg)$

Hence,

$\boxed{ \rm{ \:\rm2 {tan}^{ - 1} \bigg \{tan \frac{ \alpha }{2}. tan \bigg( \frac{\pi}{4} - \frac{ \beta }{2} \bigg) \bigg \} = {tan}^{ - 1} \bigg( \dfrac{sin \alpha \: cos \beta }{cos \alpha + sin \beta } \bigg)}}$