First slide

Extreme values and Periodicity of Trigonometric functions

Question

$In ∆ A B C, A is obtuse then$

Moderate

A

$\tan B \tan C = 1$
B

$\tan B \tan C < 1$
C

$\tan B \tan C > 1$
D

$\tan B \tan C = - 1$

Solution

A is obtuse . $\Rightarrow B + C < 90^{°} \Rightarrow B < 90^{°} - C \Rightarrow \tan B < \tan (90^{°} - C) = c o t C$ . $= \frac{1}{\tan C}$
$\Rightarrow$ $\tan B \tan C < 1$

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