plz solve this one

also,plz show me the steps of how you got the desired answer.

The answer is 3.

Sorry ... I could have helped u but I don't know how to use LATEX

Can anyone show me how the answer is c)3?

Hint: Use tan (A+B) on tan(θ+60⁰) and tan (A-B) on tan(θ-+60⁰)

Following up on the hint,
\tan (\theta - 60^{\circ}) + \tan (\theta + 60^{\circ}) = \frac{\tan \theta - \sqrt 3}{1+\sqrt 3 \tan \theta} + \frac{\tan \theta + \sqrt 3}{1-\sqrt 3 \tan \theta}

= \frac{ 8 \tan \theta}{1-3 \tan^2 \theta}

Hence the given expression

=\tan \theta + \frac{ 8 \tan \theta}{1-3 \tan^2 \theta} = \frac{ 9 \tan \theta - 3 \tan^3 \theta}{1-3 \tan^2 \theta}

=\tan \theta + \frac{ 8 \tan \theta}{1-3 \tan^2 \theta} = \frac{ 9 \tan \theta - 3 \tan^3 \theta}{1-3 \tan^2 \theta} = \frac{3(3 \tan \theta - \tan^3 \theta)}{1-3 \tan^2 \theta} = 3 \tan 3 \theta

Thank you,prophet for yur answer but i had already derived the proof after i saw the hint which you gave me....