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If $\sec\theta=m$ and $\tan\theta=n$ then $\large\frac{1}{m}$$\big[(m+n)+\large\frac{1}{(m+n)}\big]$

$\begin{array}{1 1}(a)\;2&(b)\;2m\\(c)\;2n&(d)\;mn\end{array}$

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1 Answer

Given that $\sec\theta=m$ and $\tan\theta=n$
$\therefore \large\frac{1}{m}$$\bigg[(m+n)+\large\frac{1}{m+n}\bigg]$
$\Rightarrow \large\frac{1}{\sec\theta}$$\big[\sec\theta+\tan\theta+\large\frac{1}{\sec\theta+\tan\theta}\big]$
$\Rightarrow \large\frac{1}{\sec\theta}$$\big[\sec\theta+\tan\theta+\sec\theta-\tan\theta]$
$\Rightarrow \large\frac{1}{\sec\theta}$$.2\sec\theta$
$\Rightarrow 2$
Hence (a) is the correct option.
answered Nov 15, 2013 by sreemathi.v
 

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