Question
Find the value of \tan (\alpha +\beta) such that \sin \alpha +\sin \beta = \dfrac{1}{4} and \cos \alpha +\cos \beta =\dfrac{1}{3}
Answer
\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}
\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}
\sin \alpha +\sin \beta =2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}= \dfrac{1}{4}
(1)
\cos \alpha +\cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}=\dfrac{1}{3}
(2)
\tan \dfrac{\alpha +\beta }{2} = \dfrac{3}{4}
(3)
Double angle identity
\tan (\alpha +\beta) = \dfrac{2\tan\dfrac{\alpha +\beta }{2} }{1-\tan^2\dfrac{\alpha +\beta }{2} }
=\dfrac{2\cdotp \dfrac{3}{4} }{1-(\dfrac{3}{4} )^2}
=\dfrac{24}{7}