If tan20o=λ,thentan250o+tan340otan200o−tan110o=

=tan(270o−20o)+tan(360o−20o)tan(180o+20o)−tan(90o+20o)=cot20o−tan20otan20o+cot20o=1/λ−λλ+1/λ=1−λ21+λ2

If tan20o=λ,thentan250o+tan340otan200o−tan110o=

=tan(270o−20o)+tan(360o−20o)tan(180o+20o)−tan(90o+20o)=cot20o−tan20otan20o+cot20o=1/λ−λλ+1/λ=1−λ21+λ2

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If $\tan 20^{o} = λ, t h e n \frac{\tan 250^{o} + \tan 340^{o}}{\tan 200^{o} - \tan 110^{o}} =$

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$\frac{1 + λ^{2}}{1 - λ^{2}}$

$\frac{1 - λ^{2}}{2 λ}$

$\frac{1 - λ^{2}}{1 + λ^{2}}$

$\frac{1 + λ^{2}}{2 λ}$

answer is B.

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Detailed Solution

$= \frac{\tan (270^{o} - 20^{o}) + \tan (360^{o} - 20^{o})}{\tan (180^{o} + 20^{o}) - \tan (90^{o} + 20^{o})} = \frac{\cot 20^{o} - \tan 20^{o}}{\tan 20^{o} + \cot 20^{o}} = \frac{1 / λ - λ}{λ + 1 / λ} = \frac{1 - λ^{2}}{1 + λ^{2}}$

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If tan20o=λ,thentan250o+tan340otan200o−tan110o=

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If $\tan 20^{o} = λ, t h e n \frac{\tan 250^{o} + \tan 340^{o}}{\tan 200^{o} - \tan 110^{o}} =$