Question
Question: If \(\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan \theta \). Then the value of \(\thet...
If 1−tan230∘2tan30∘=tanθ. Then the value of θ is:
(a) 60∘
(b) 65∘
(c) 45∘
(d) 30∘
Solution
Hint: Try to simplify the left-hand side of the equation that is given in the question by using the formula tan2A=1−tan2A2tanA, and other similar formulas.
Complete step-by-step answer:
We will now solve the left-hand side of the equation given in the question.
1−tan230∘2tan30∘
Now we know that tan2A=1−tan2A2tanA . Therefore, our equation becomes:
tan(2×30)∘
=tan60∘
Therefore, we can say that:
tanθ=tan60∘
Now according to the rule of trigonometric equations: tanβ=tanθ implies that θ=n×180∘+β .
∴θ=n×180∘+60∘
Now, as n can be any integer, we let n=0.
θ=60∘
Hence, the answer to the above question is option (a).
Note: Be careful, as 60∘ is not the only possible value of θ . If we substitute the value of n in θ=n×180∘+60∘ with different integers, the different values of θ we get are also acceptable values of θ . Also, we need to remember the properties related to complementary angles and trigonometric ratios.