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Q.

If the set of values of x satisfying the inequality tanx. Tan3x<-1 in the interval (0, $\frac{π}{2}$ ) is (a, b) then the value of $(\frac{36 (b - a)}{π})$ is

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answer is 3.

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Detailed Solution

$\tan x \tan 3 x < - 1 \tan (\frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x}) <$
Let tan x = t
$\Rightarrow \frac{(1 - t) (1 + t)}{1 - 3 t^{2}} < 0 \Rightarrow t \in (- 1, \frac{- 1}{\sqrt{3}})$
But only $\tan x \in (\frac{1}{\sqrt{3}}, 1)$
$\Rightarrow x \in (\frac{π}{6}, \frac{π}{4}) \Rightarrow a = \Rightarrow \frac{36}{π} (b - a) = 3$

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