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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the value of $\sec(-\large\frac{25\pi}{3})$

$\begin{array}{1 1}(A)\;2&(B)\;4\\(C)\;3&(D)\;8\end{array} $

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1 Answer

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  • $\sec(\theta+2n\pi)=\sec\theta$
$\sec(\large\frac{-25\pi}{3})$$=\sec(\large\frac{25\pi}{3})$
$\cos(-\theta)=\cos \theta$
$\sec(8\pi+\large\frac{\pi}{3})$
$\sec\large\frac{\pi}{3}$$=2$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v
 
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