logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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.
answered May 22, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...