Ask Questions, Get Answers

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

The value of $\tan 3A-\tan 2A-\tan A$ is equal to

$\begin{array}{1 1}(A)\;\tan 3A\tan 2A\tan A\\(B)\;-\tan 3A\tan 2A\tan A\\(C)\;\tan A\tan 2A-\tan 2A\tan 3A-\tan 3A\tan A\\(D)\;\text{None of these}\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $\tan 3A=\large\frac{3\tan A-\tan^3A}{1-3\tan^2A}$
  • $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan 3A-\tan 2A-\tan A$
$\tan 3A-\tan A-\tan 2A$
$\large\frac{3\tan A-\tan^3A}{1-3\tan^2A}$$-\tan A-\large\frac{2\tan A}{1-\tan^2A}$
$\large\frac{\tan A(3-\tan^2A-1+3\tan^2A)}{1-3\tan^2A}-\frac{2\tan A}{1-\tan^2A}$
$2\tan A\large\frac{(1+\tan^2A}{1-3\tan^2A}-\frac{2\tan A}{1-\tan^2A}$
$\large\frac{2\tan A}{1-\tan^2A}\big(\large\frac{3\tan A-\tan^3A}{1-3\tan^2A}\big)$$\tan A$
$\tan 2A.\tan 3A.\tan A$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v

Related questions

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