Question

sinA+sinB /cos A +cos B = tan (A+B/2)prove it
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Answer 1

Question :-

Prove it :

$\tt \frac{ \sin A+ \sin B }{ \cos A + \cos B} = \tan \bigg(\frac{A+ B }{2} \bigg)$

Used Formula :-

$\star \: \tt\sin C + \sin D = 2 \sin \bigg( \frac{ C + D }{2} \bigg) . \cos \: \bigg( \frac{ C - D }{2} \bigg) \\ \\ \star \: \tt\cos C + \cos D = 2 \cos \bigg( \frac{ C + D }{2} \bigg) . \cos \: \bigg( \frac{ C - D }{2} \bigg)$

Explanation :-

Taking left hand side

$\to \: \tt \frac{ \sin A+ \sin B }{ \cos A + \cos B} \: \\ \\ \sf \: apply \: the \: given \: formula \ \\ \\ \to \tt\frac{ \cancel{2} \sin \bigg( \frac{ A+ B }{2} \bigg) . \cancel{ \cos \: \bigg( \frac{A - B }{2} \bigg)}}{ \cancel{2} \cos\bigg( \frac{ A+ B }{2} \bigg) . \cancel{ \cos \: \bigg( \frac{ A - B }{2} \bigg)} \: } \\ \\ \to \tt \frac{\sin \bigg( \frac{ A+ B }{2} \bigg) }{\cos \bigg( \frac{ A+ B }{2} \bigg)} \\ \\ \boxed{ \tt \because \tan \theta = \frac{ \sin \theta}{ \cos \theta} } \\ \\ \to \: \tan \bigg( \frac{ A+ B }{2} \bigg) \:$

→ LHS = RHS

Hence proved .