Question

Prove that :
cot a + cot (60° + a) - cot (60° -a) = 3 cot 3a​

Answer 1

Solution :

$\sf \cot a + \cot(60 + a) - \cot(60 - a) = 3 \cot3a$

On solving LHS

$\implies \sf\cot a + \cot(60 + a) - \cot(60 - a) \\ \\ \implies\sf \cot a + \frac{ \sin(60 + a) }{ \cos(60 + a) } - \frac{ \cos(60 - a) }{ \sin(60 - a) } \\ \\ \implies \sf \cot a + \frac{ \cos a - \sqrt{3} \sin a }{ \sqrt{3} \cos a + \sin a} - \frac{ \cos a + \sqrt{3} \sin a }{ \sqrt{3} \cos a - \sin a } \\ \\ \implies \sf \cot a + \frac{ - 8 \sin a \cos a}{ 3 { \cos}^{2}a - {sin}^{2} a } \\ \\ \implies \sf \cot a - \frac{8 \cot a}{3 { \cot}^{2}a - 1 } \\ \\ \implies\sf \frac{3{ \cot}^{3} a - 9 \cot a}{3 { \cot}^{2} a - 1} \\ \\ \implies \sf 3 \bigg( \frac{ { \cot}^{3}a - 3 \cot a }{3 {cot}^{2}a - 1 } \bigg ) \\ \\ \implies \sf 3 \cot3a$

Which is equals to RHS

Hence Proved

Important trigonometric identities :

$\sf \sin(a+b) = \sin A \cos B+ \cos A\sin B\\ \\ \cos(a+b) = \cos A \cos B - \sin A \sin B \\ \\ \tan(A+B) = \frac{\tan A \tan B}{1-\tan A \tan B}$