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Evaluate $\tan 75^{\large\circ}+\cot 75^{\large\circ}$

$\begin{array}{1 1}(A)\;2&(B)\;4\\(C)\;6&(D)\;8\end{array} $

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  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
  • $\cot \theta=\large\frac{1}{\tan \theta}$
$\tan 75^{\large\circ}=\tan (45^{\large\circ}+30^{\large\circ})$
$\Rightarrow \large\frac{\tan 45^{\large\circ}+\tan 30^{\large\circ}}{1-\tan 45^{\large\circ}\tan 30^{\large\circ}}$
$\Rightarrow \large\frac{1+\Large\frac{1}{\sqrt 3}}{1-\Large\frac{1}{\sqrt 3}}$
$\Rightarrow \large\frac{\sqrt 3+1}{\sqrt 3-1}$
$\cot 75^{\large\circ}=\large\frac{1}{\tan 75^{\large\circ}}$
$\Rightarrow \large\frac{\sqrt 3-1}{\sqrt 3+1}$
$\tan 75^{\large\circ}+\cot 75^{\large\circ}=\large\frac{\sqrt 3+1}{\sqrt 3-1}+\large\frac{\sqrt 3-1}{\sqrt 3+1}$
$\Rightarrow \large\frac{(\sqrt 3+1)^2+(\sqrt 3-1)^2}{(\sqrt 3-1)(\sqrt 3+1)}$
$\Rightarrow \large\frac{(4+2\sqrt 3)+(4-2\sqrt 3)}{3-1}$
$\Rightarrow \large\frac{8}{2}$
$\Rightarrow 4$
Hence (B) is the correct answer.
answered May 21, 2014 by sreemathi.v
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