If Tan40∘+2 Tan10∘=Cotx then x=

tan400+2tan400=cot xtan(500-400)=tan100tan500-tan4001+tan500tan400=tan100tan500-tan4001+1=tan100tan500-tan400=2tan100tan400+2tan100=tan500 =cot400

If Tan40∘+2 Tan10∘=Cotx then x=

tan400+2tan400=cot xtan(500-400)=tan100tan500-tan4001+tan500tan400=tan100tan500-tan4001+1=tan100tan500-tan400=2tan100tan400+2tan100=tan500 =cot400

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If $Tan 40^{\circ} + 2 Tan 10^{\circ} = Cotx$ then $x =$

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$75^{\circ}$

$85^{\circ}$

$30^{\circ}$

$40^{\circ}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \tan 40^{0} + 2 \tan^{40^{0}} & = & \cot x \\ \tan (50^{0} - 40^{0}) & = & \tan 10^{0} \\ \frac{\tan 50^{0} - \tan 40^{0}}{1 + \tan 50^{0} \tan 40^{0}} & = & \tan 10^{0} \\ \frac{\tan 50^{0} - \tan 40^{0}}{1 + 1} & = & \tan 10^{0} \\ \tan 50^{0} - \tan 40^{0} & = & 2 \tan 10^{0} \\ \tan 40^{0} + 2 \tan 10^{0} & = & \tan 50^{0} \\ = & \cot 40^{0} \end{array}$