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

The correct option is D π/4 Given that: tanα =x/x+1 and tanβ =1/2x+1 Now, tan(α +β )=tanα +tanβ/1 tanα +tanβ ⇒ tan(α +β )=x/x+1+1/2x+1/1 ( x/x+1 ) ( x/2x+1 ) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tanα=x/x+1...

Question

If $t a n α = \frac{x}{x + 1} a n d t a n β = \frac{1}{2 x + 1}$ , then $α + β$ is equal to

$\frac{π}{2}$

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$\frac{π}{3}$

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$\frac{π}{6}$

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$\frac{π}{4}$

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Solution

The correct option is D
$\frac{π}{4}$

Given that:
$t a n α = \frac{x}{x + 1} a n d t a n β = \frac{1}{2 x + 1}$
Now, $t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α + t a n β} \Rightarrow t a n (α + β) = \frac{\frac{x}{x + 1} + \frac{1}{2 x + 1}}{1 - (\frac{x}{x + 1}) (\frac{x}{2 x + 1})} \Rightarrow t a n (α - β) = \frac{x (2 x + 1) + x + 1}{(x + 1) (2 x + 1) - 2 x} \Rightarrow t a n (α + - β) = \frac{2 x^{2} + x + x + 1}{2 x^{2} + 2 x + x + 1 - x} \Rightarrow t a n (α + β) = \frac{2 x^{2} + x + x + 1}{2 x^{2} + 2 x + 1} \Rightarrow t a n (α + β) = 1 ∴ α + β = \frac{π}{4}$

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If tanα=xx+1 and tanβ=12x+1, then α+β is equal to

The correct option is D π4 Given that: tanα=xx+1 and tanβ=12x+1 Now, tan(α+β)=tanα+tanβ1−tanα+tanβ⇒tan(α+β)=xx+1+12x+11−(xx+1)(x2x+1)⇒tan(α−β)=x(2x+1)+x+1(x+1)(2x+1)−2x⇒tan(α+−β)=2x2+x+x+12x2+2x+x+1−x⇒tan(α+β)=2x2+x+x+12x2+2x+1⇒tan(α+β)=1∴ α+β=π4

If $t a n α = \frac{x}{x + 1} a n d t a n β = \frac{1}{2 x + 1}$ , then $α + β$ is equal to