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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the value : $\tan 15^{\large\circ}$

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

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  • $\tan (A-B)=\large\frac{\tan A-\tan B}{1+\tan A \tan B}$
$\tan 15^{\large\circ}=\tan(45^{\large\circ}-30^{\large\circ})$
$\Rightarrow \large\frac{\tan^{\large\circ}-\tan 30^{\large\circ}}{1+\tan 45^{\large\circ}.\tan 30^{\large\circ}}$
$\Rightarrow \large\frac{1-\large\frac{1}{\sqrt 3}}{1+1\times \large\frac{1}{\sqrt 3}}=\large\frac{\Large\frac{\sqrt 3-1}{\sqrt 3}}{\Large\frac{\sqrt 3-1}{\sqrt 3}}$
$\Rightarrow \large\frac{\sqrt 3-1}{\sqrt 3+1}$
Multiplying both numerator and denominator by $\sqrt 3-1$
$\large\frac{\sqrt 3-1}{\sqrt 3+1}\times \frac{\sqrt 3-1}{\sqrt 3-1}=\frac{(\sqrt 3-1)^2}{(\sqrt 3)^2-1^2}$
$\Rightarrow \large\frac{(\sqrt 3)^2-2\sqrt 3+1^2}{3-1}$
$\Rightarrow \large\frac{3-2\sqrt 3+1}{2}$
$\Rightarrow \large\frac{4-2\sqrt 3}{2}$
$\Rightarrow \large\frac{2(2-\sqrt 3)}{2}$
$\Rightarrow 2-\sqrt 3$
Hence (A) is the correct answer.
answered Apr 17, 2014 by sreemathi.v

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