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${(\frac{\tan 70^{o} - \tan 20^{o}}{2 \tan 50^{o}} - \frac{\tan 50^{o} - \tan 40^{o}}{2 \tan 10^{o}})}^{2020} =$

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Correct option is A

If A+B=90°, then tanA-tanB=2tan(A-B)∴tan70°-tan20°2tan50°=2tan50°2tan50°=1 and tan50°-tan40°2tan10°=2tan10°2tan10°=1 ∴tan70°-tan20°2tan50°-tan50°-tan40°2tan10°2020=1-12020=0

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tan70o−tan20o2tan50o−tan50o−tan40o2tan10o2020=

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${(\frac{\tan 70^{o} - \tan 20^{o}}{2 \tan 50^{o}} - \frac{\tan 50^{o} - \tan 40^{o}}{2 \tan 10^{o}})}^{2020} =$