If A, B, C are the angles of a triangle such that C is an obtuse angle, then

t a n A . t a n B < 1

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t a n A . t a n B > 1

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t a n A . t a n B = 1

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None

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Solution

The correct option is A $t a n A . t a n B < 1$
Since C is obtuse therefore A and B both are acute and A + B is also acute.
Hence $t a n A, t a n B$ and $t a n (A + B)$ are all + ive
$t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B} = + i v e$
Above is possible only when $t a n A t a n B < 1$

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