A) \[2-\sqrt{3}\]
B) \[\frac{2}{\sqrt{3}}\]
C) \[\sqrt{3}-\sqrt{2}\]
D) \[4-2\sqrt{3}\]
Correct Answer: D
Solution :
\[A,\,B\,>\,0\]and A \[A+B=\frac{\pi }{6}\] Let y = tanA + tanB \[\frac{dy}{dA}={{\sec }^{2}}A-{{\sec }^{2}}\left( \frac{\pi }{6}-A \right)\] Hence\[\tan A+\tan B\uparrow \forall A\in \left[ \frac{\pi }{12},\frac{\pi }{6} \right]\] and \[\tan A+\tan B\downarrow \forall A\in \left[ 0,\frac{\pi }{12} \right]\], clearly tanA + tanB is minimum when \[A=B=\frac{\pi }{12}\] \[\Rightarrow \,{{y}_{\min }}=2\tan \frac{\pi }{12}\] \[=\left( 2-\sqrt{3} \right)\times 2=4-2\sqrt{3}\]
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