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$if( \alpha + \beta )is \: not \: equel \: to \: zero(0) \: \:and \: \\ \sin \alpha + \sin \beta = 2 \sin( \alpha + \beta ) \: then \: prove \: that \\ \tan \frac{ \alpha }{2} \tan \frac{ \beta }{2} = \frac{1}{3}$







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Answer 1

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given \: that : \\ \\ \alpha + \beta \neq \: 0 \\ \\ \sf \: also \: given : \\ \\ \rm \: sin \: \alpha + sin \: \beta = 2 \: sin( \alpha + \beta ) \\ \\ \red{ \underline{ \bf \: we \: have \: to \: prove : }} \\ \\ \rm \: tan \: \frac{ \alpha }{2}.tan \: \frac{ \beta }{2} = \frac{1}{3} \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc} \red { \underline{\bf \: solution}} \\ \\ \rm \: sin \: \alpha + \: sin \: \beta = 2 \: sin( \alpha + \beta ) \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know : \\ \\ \to \: \bf \: sin \: x + sin \: y = 2 \: sin \: ( \frac{x + y}{2}) . \: cos \: ( \frac{x - y}{2} ) \\ \\ \rm \: \to \: sin \: 2x = 2 \: sin \: x \: cos \: x \\ \therefore \: \rm \: sin \: x = 2 \: sin \: \frac{x}{2} . \: cos \: \frac{x}{2} \end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ \rm \implies \: \cancel2 \: sin \: ( \frac{ \alpha + \beta }{2} ). cos \: ( \frac{ \alpha - \beta }{2}) = \cancel 2 \times 2 \: sin ( \frac{ \alpha + \beta }{2} ).cos( \frac{ \alpha + \beta }{2} ) \\ \\ \rm \implies \:sin( \frac{ \alpha + \beta }{2} ) \{ cos( \frac{ \alpha - \beta }{2} )- 2 \: cos( \frac{ \alpha + \beta }{2} ) \} = 0 \\ \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc}\sf \: since \: \: \alpha + \beta \neq \: 0 \: \: \: sin( \frac{ \alpha + \beta }{2}) \neq0 \\ \\ \\ \therefore \: \rm \: cos( \frac{ \alpha - \beta }{2} ) - 2 \: cos( \frac{ \alpha + \beta }{2} ) = 0 \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know: \\ \\ \rm \: \to \: cos(x \pm y) = cos \: x \: cos \: y \mp \: sin \: x \: sin \: y\end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ \rm \implies \: cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} + sin \frac{ \alpha }{2} .sin \frac{ \beta }{2} - 2(cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} - sin \frac{ \alpha }{2}.sin \frac{ \beta }{2} ) = 0 \\ \\ \rm \implies \: 3 \: sin \frac{ \alpha }{2} \: sin \frac{ \beta }{2} = cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} \\ \\ \rm \implies \: \frac{sin \frac{ \alpha }{2} .sin \frac{ \beta }{2} }{cos \frac{ \alpha }{2}.cos \frac{ \beta }{2} } = \frac{1}{3} \\ \\ \rm \implies \: tan \: \frac{ \alpha }{2} .tan \frac{ \beta }{2} = \frac{1}{3} \\ \\ \therefore \: L.H.S = R.H.S\end{array}}}}$