Question

(cos 7A + cos 5A)÷(sin 7A + sin 5A)=cot 6A.​

Answer 1

$\maltese \: \: \: \: \textbf{\underline{\underline{Solution \: : }}}$

$\sf \longrightarrow \: \frac{ \cos(7A) + \cos(5A) }{ \sin(7A) + \sin(5A) }$

$\sf \longrightarrow \: \frac{ 2 \cos( \frac{7A + 5A}{2}) \cos( \frac{7A - 5A}{2}) }{2 \sin( \frac{7A + 5A}{2}) \cos( \frac{7A - 5A}{2} ) }$

$\sf \longrightarrow \: \frac{ \cancel{2 } \: \cos( \frac{7A + 5A}{2}) \: \cancel{\cos( \frac{7A - 5A}{2})} }{ \cancel{2} \: \sin( \frac{7A + 5A}{2}) \: \cancel{\cos( \frac{7A - 5A}{2} ) }}$

$\longrightarrow \: { \underline{\boxed{ \bf{\cot(6A) }} \: \: }}_{ \bigstar \star} \: \: \: \: \: \: \: \: \bf{[Proved]}$

$\maltese \: \: \: \: \textbf{\underline{\underline{Required \: Concept \: : }}}$

${ \large{\odot }}\: \: \: \bf \: \sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2}$

${ \large{ \odot }}\: \: \: \bf \: \cos C + \cos D = 2 \cos \frac{C + D}{2} \cos \frac{C - D}{2}$