If tan⁡α=xx+1 and tan⁡β=12x+1, then α+β

tan(α+β)=tanα+tanβ1-tanαtanβtan(α+β)=xx+1+12x+11-xx+112x+1Will get, tan(α+β)=1 then (α+β)=π4

If tan⁡α=xx+1 and tan⁡β=12x+1, then α+β

tan(α+β)=tanα+tanβ1-tanαtanβtan(α+β)=xx+1+12x+11-xx+112x+1Will get, tan(α+β)=1 then (α+β)=π4

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$If \tan α = \frac{x}{x + 1} and \tan β = \frac{1}{2 x + 1}, then α + β$

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$π / 2$

$π / 4$

$π / 3$

answer is B.

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$\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$

$\tan (α + β) = \frac{\frac{x}{x + 1} + \frac{1}{2 x + 1}}{1 - (\frac{x}{x + 1}) (\frac{1}{2 x + 1})}$

Will get, $\tan (α + β) = 1$ then $(α + β) = \frac{π}{4}$

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If tan⁡α=xx+1 and tan⁡β=12x+1, then α+β

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$If \tan α = \frac{x}{x + 1} and \tan β = \frac{1}{2 x + 1}, then α + β$