Ask Questions, Get Answers

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

True-or-False :If $\tan A=\large\frac{1-\cos B}{\sin B}$ then $\tan 2A=\tan B$

$\begin{array}{1 1}(A)\;\text{True}\\(B)\;\text{False}\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\Rightarrow \large\frac{2(\Large\frac{1-\cos B}{\sin B}}{1-\Large\frac{(1-\cos B)^2}{\sin ^2B}}$
$\Rightarrow \large\frac{2(1-\cos B)}{\sin B}\times \frac{\sin ^2B}{\sin ^2B-(1-\cos B)^2}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{\sin ^2B-(1+\cos ^2B-2\cos B)}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{\sin ^2B-1-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{-\cos ^2B-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{-2\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{2\cos B(1-\cos B)}$
$\Rightarrow \tan B$
Hence the given statement is true.
answered May 20, 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