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The expression $\tan^2\alpha+\cot^2\alpha$ is

$(a)\;\geq 2\qquad(b)\;\leq 2\qquad(c)\;\geq -2\qquad(d)\;None\;of\;these$

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1 Answer

$(\tan\alpha-\cot\alpha)^2\geq 0$
$\Rightarrow \tan^2\alpha+\cot^2\alpha-2\tan\alpha\cot\alpha\geq 0$
$\Rightarrow \tan^2\alpha+\cot^2\alpha-2\geq 0$
$\Rightarrow \tan^2\alpha+\cot^2\alpha\geq 2$
Hence (a) is the correct answer.
answered Oct 9, 2013 by sreemathi.v
 

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