If tan-x-x-1 and tan-1-2x-1-then - - is -

The correct option is C π/4 We know, tan(α+β) #160; =tanα+tanβ1 tanα.tanβ =x(2x+1)+x+1(2x+1)(x+1) x =2x^2+2x+12x^2+2x+x+1 x =2x ...

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If tanα=x/x...

Question

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

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

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Similar questions

t a n α = \frac{x}{x + 1}

t a n β = \frac{1}{2 x + 1},

α + β

I f t a n α = \frac{m}{(m + 1)}, t a n β = \frac{1}{(2 m + 1)}, t h e n α + β = (n ϵ Z)

tan α = \frac{1}{2}, tan β = \frac{1}{3}

α + β = π / 4

0 < α < β < \frac{π}{2}

\frac{tan β}{tan α} > \frac{α}{β}

x tan x

0 < x < \frac{π}{2}

t a n α = \frac{1}{1 + 2^{- x}}

t a n β = \frac{1}{1 + 2^{x + 1}}

α + β

(0, \frac{π}{2})

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