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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

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} $

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1 Answer

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  • $\tan 3A=\large\frac{3\tan A-\tan^3A}{1-3\tan^2A}$
  • $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan 3A-\tan 2A-\tan A$
$\tan 3A-\tan A-\tan 2A$
$\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
 

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