Question

Tan 60°= 2 tan 30°/1- tan2 30°

Answer 1

Question :-

Prove that

$\tan 60 \degree = \frac{2 \tan 30 \degree}{1 - { \tan}^{2} 30 \degree}$

Proof :-

Taking left hand side

→ LHS = tan 60° = √3

now taking right hand side

$\implies \: \frac{2 \tan 30 \degree}{1 - { \tan}^{2} 30 \degree} \because \tan 30 = \frac{1}{ \sqrt{3} } \: \\ \\ \implies \: \frac{2 \frac{1}{ \sqrt{3} } }{1 - { \big( \frac{1}{ \sqrt{3} } \big) }^{2} } \\ \\ \implies \: \frac{ \frac{2}{ \sqrt{3} } }{ 1 - \frac{1}{3} } \\ \\ \implies \: \frac{ \frac{2}{ \sqrt{3} } }{ \frac{2}{3} } \\ \\ \implies \: \frac{3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \implies \: \frac{3 \sqrt{3} }{3} = \sqrt{3}$

LHS = RHS

hence proved .

Answer 2

$\huge\tt{Solution:-}$

$tan60° = \frac{2\:tan30° }{1 - {tan}^{2}30°} \\ \\ LHS = \frac{2tan30° }{1 - {tan}^{2}30°} \\ \\ LHS. = \frac{2 × \frac{1}{\sqrt{3}}}{1 - {{\frac{1}{\sqrt{3}}}}^{2}} \\ \\ LHS = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}}\\ \\LHS= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \\ \\LHS= \frac{3}{\sqrt{3}} \\ \\ LHS = \frac{3}{\sqrt{3}} × \frac{\sqrt{3}}{\sqrt{3}}\\ \\ LHS = \frac{3\sqrt{3}}{3} \\ \\ LHS = \sqrt{3} \\ \\ LHS = tan60° \\ \\ HENCE \: PROVED$