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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

Solve the equation $\tan\theta+\tan(\theta+\large\frac{\pi}{3})$$+\tan(\theta+\large\frac{2\pi}{3})=$$3$

$\begin{array}{1 1}(A)\;\tan\large\frac{\pi}{4}&(B)\;\tan \pi\\(C)\;\tan\large\frac{\pi}{2}&(D)\;\tan 2\pi\end{array} $

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A)
Toolbox:
  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
Given
$\tan \theta+\tan(\theta+\large\frac{\pi}{3})+$$\tan (\theta+\large\frac{2\pi}{3})$$=3$
$\tan \theta+\large\frac{\tan \theta+\tan\large\frac{\pi}{3}}{1-\tan \theta\tan \large\frac{\pi}{3}}+\large\frac{\tan \theta+\tan\large\frac{2\pi}{3}}{1-\tan \theta\tan \large\frac{2\pi}{3}}$$=3$
$\tan\theta+\large\frac{\tan \theta+\sqrt 3}{1-\sqrt 3\tan \theta}+\large\frac{\tan \theta-\sqrt 3}{1+\sqrt 3\tan \theta}=$$=3$
$\tan \theta+\large\frac{8\tan \theta}{1-3\tan^2\theta}$$=3$
$\Rightarrow \large\frac{\tan \theta-3\tan^3\theta+8\tan \theta}{1-3\tan^2\theta}$$=3$
$3\tan 3\theta=3$
$\tan 3\theta=1$
$\tan 3\theta=\tan \large\frac{\pi}{4}$
Hence (A) is the correct answer.
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