1The answer is 3.
Sorry ... I could have helped u but I don't know how to use LATEX
tanθ+tan(θ+60)+tan(θ-60)=k tan 3θ,where k is equal to:
a)1
b)2
c)3
d)4
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10 Answers
1
341Hint: Use tan (A+B) on tan(θ+600) and tan (A-B) on tan(θ-+600)
1just use the identity tan(a+b) .open it and use lcm u will get answer
341Following 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
1Thank you,prophet for yur answer but i had already derived the proof after i saw the hint which you gave me....