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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $A+B=\large\frac{\pi}{3}$ where $A,B\in R^+$ then the minimum value of $\sec A+\sec B$ is equal to

$(a)\;\large\frac{2}{\sqrt 3}\qquad$$(b)\;\large\frac{4}{\sqrt 3}$$\qquad(c)\;2\sqrt 3\qquad(d)\;None\;of\;these$

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For $y=\sec x,x\in [0,\large\frac{\pi}{2}]$ tangent drawn to it any point lies completely below the graph of $y=\sec x$ thus
$\large\frac{\sec A\sec B}{2}$$\geq \sec\big(\large\frac{A+B}{2}\big)$
$\Rightarrow \sec A+\sec B\geq 2\sec\big(\large\frac{\pi}{6}\big)$
$\Rightarrow \large\frac{4}{\sqrt 3}$
Hence (b) is right option.
answered Oct 7, 2013 by sreemathi.v

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