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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The value of $\cot(\large\frac{\pi}{4}+$$\theta).\cot(\large\frac{\pi}{4}$$-\theta)$ is

$\begin{array}{1 1}(A)\;-1&(B)\;0\\(C)\;1&(D)\;\text{Not defined}\end{array} $

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1 Answer

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Toolbox:
  • $\cot(a+b)=\large\frac{1-\tan a\tan b}{\tan a+\tan b}$
  • $\cot(a-b)=\large\frac{1+\tan a\tan b}{\tan a-\tan b}$
$\cot(\large\frac{\pi}{4}+$$\theta).\cot(\large\frac{\pi}{4}$$-\theta)$
$\large\frac{1-\tan \large\frac{\pi}{4}\normalsize\tan\theta}{\tan\large\frac{\pi}{4}+\normalsize \tan\theta}\times \large\frac{1+\tan \large\frac{\pi}{4}\normalsize\tan\theta}{\tan\large\frac{\pi}{4}-\normalsize \tan\theta}$
$\large\frac{1-\tan \theta}{1+\tan \theta}\times \large\frac{1+\tan \theta}{1-\tan \theta}$$=1$
Hence (C) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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