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Question

Question: If \(\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan \theta \). Then the value of \(\thet...

If 2tan301tan230=tanθ\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan \theta . Then the value of θ\theta is:
(a) 6060{}^\circ
(b) 6565{}^\circ
(c) 4545{}^\circ
(d) 3030{}^\circ

Explanation

Solution

Hint: Try to simplify the left-hand side of the equation that is given in the question by using the formula tan2A=2tanA1tan2A\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}, and other similar formulas.

Complete step-by-step answer:
We will now solve the left-hand side of the equation given in the question.
2tan301tan230\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }
Now we know that tan2A=2tanA1tan2A\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A} . Therefore, our equation becomes:
tan(2×30)\tan \left( 2\times 30 \right){}^\circ
=tan60=\tan 60{}^\circ
Therefore, we can say that:
tanθ=tan60\tan \theta =\tan 60{}^\circ
Now according to the rule of trigonometric equations: tanβ=tanθ\tan \beta =\tan \theta implies that θ=n×180+β\theta =n\times 180{}^\circ +\beta .
θ=n×180+60\therefore \theta =n\times 180{}^\circ +60{}^\circ
Now, as n can be any integer, we let n=0.
θ=60\theta =60{}^\circ
Hence, the answer to the above question is option (a).

Note: Be careful, as 6060{}^\circ is not the only possible value of θ\theta . If we substitute the value of n in θ=n×180+60\theta =n\times 180{}^\circ +60{}^\circ with different integers, the different values of θ\theta we get are also acceptable values of θ\theta . Also, we need to remember the properties related to complementary angles and trigonometric ratios.