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 $\tan 75^{\large\circ}-\cot 75^{\large\circ}$ is equal to

$\begin{array}{1 1}(A)\;2\sqrt 3&(B)\;2+\sqrt 3\\(C)\;2-\sqrt 3&(D)\;1\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $\tan(A+B)=\large\frac{\tan a+\tan B}{1-\tan A\tan B}$
$\tan 75^{\large\circ}-\cot 75^{\large\circ}$
$\tan 75^{\large\circ}-\large\frac{1}{\tan 75^{\large\circ}}$
$\tan (45^{\large\circ}+30^{\large\circ})-\large\frac{1}{\tan( 45^{\large\circ}+30^{\large\circ})}$
$\large\frac{\tan 45^{\large\circ}+\tan 30^{\large\circ}}{1-\tan 45^{\large\circ}\tan 30^{\large\circ}}-\frac{1-\tan 45^{\large\circ}\tan 30^{\large\circ}}{\tan 45^{\large\circ}\tan 30^{\large\circ}}$
$\large\frac{1+\Large\frac{1}{\sqrt 3}}{1-\Large\frac{1}{\sqrt 3}}-\large\frac{1-\Large\frac{1}{\sqrt 3}}{1+\Large\frac{1}{\sqrt 3}}$
$\large\frac{(1+\Large\frac{1}{\sqrt 3})^2-(1+\Large\frac{1}{\sqrt 3})^2}{1-\Large\frac{1}{3}}$
$\large\frac{1+\large\frac{1}{3}+\frac{2}{\sqrt 3}\normalsize -1-\large\frac{1}{3}+\frac{2}{\sqrt 3}}{\Large\frac{2}{3}}$
$\large\frac{4}{\sqrt 3}\times \frac{3}{2}$
$2\sqrt 3$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App