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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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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} $

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1 Answer

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  • $\tan(A+B)=\large\frac{\tan a+\tan B}{1-\tan A\tan B}$
$\tan 75^{\large\circ}-\cot 75^{\large\circ}$
$\tan 75^{\large\circ}-\large\frac{1}{\tan 75^{\large\circ}}$
$\tan (45^{\large\circ}+30^{\large\circ})-\large\frac{1}{\tan( 45^{\large\circ}+30^{\large\circ})}$
$\large\frac{\tan 45^{\large\circ}+\tan 30^{\large\circ}}{1-\tan 45^{\large\circ}\tan 30^{\large\circ}}-\frac{1-\tan 45^{\large\circ}\tan 30^{\large\circ}}{\tan 45^{\large\circ}\tan 30^{\large\circ}}$
$\large\frac{1+\Large\frac{1}{\sqrt 3}}{1-\Large\frac{1}{\sqrt 3}}-\large\frac{1-\Large\frac{1}{\sqrt 3}}{1+\Large\frac{1}{\sqrt 3}}$
$\large\frac{(1+\Large\frac{1}{\sqrt 3})^2-(1+\Large\frac{1}{\sqrt 3})^2}{1-\Large\frac{1}{3}}$
$\large\frac{1+\large\frac{1}{3}+\frac{2}{\sqrt 3}\normalsize -1-\large\frac{1}{3}+\frac{2}{\sqrt 3}}{\Large\frac{2}{3}}$
$\large\frac{4}{\sqrt 3}\times \frac{3}{2}$
$2\sqrt 3$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
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