If $Tan 70 ° - Tan 20 ° = k Cot 40 °$ then $k =$

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$- 2$

$- 1$

answer is D.

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Detailed Solution

$\tan (70^{0} - 20^{0}) = \tan 50^{0} \frac{\tan 70^{0} - \tan 20^{0}}{1 + \tan 70^{0} \cdot \tan 20^{0}} = \tan 50^{0} \frac{\tan 70^{0} - \tan 20^{0}}{1 + 1} = \tan (90^{0} - 40^{0}) \tan 70^{0} - \tan 20^{0} = 2 \cot 40^{0} k = 2$

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