If tan-x-x-1 and tan-1-2 x-1- then - is equal to-A-B- -3C- -6D- -4

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

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If tanα=x/x+1...

Question

If $t a n α = \frac{x}{x + 1}$ and $t a n β = \frac{1}{2 x + 1},$ 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,
$t a n α = \frac{x}{x + 1}$ and $t a n β = \frac{1}{2 x + 1}$
Now, $t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α . t a n β}$
$\Rightarrow t a n (α + β) = \frac{\frac{x}{x + 1}}{\frac{1}{2 x + 1}}$
$\Rightarrow t a n (α + β) = \frac{x (2 x + 1) + x + 1}{(x + 1) (2 x + 1) - x}$
$\Rightarrow t a n (α + β) = \frac{2 x^{2} + x + x + 1}{2 x^{2} + 2 x + x + 1 - x}$
$\Rightarrow t a n (α + β) = \frac{2 x^{2} + 2 x + 1}{2 x^{2} + 2 x + 1}$
$\Rightarrow t a n (α + β) = 1$
$∴ α + β = \frac{π}{4}$

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Similar questions

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{m}{m + 1}

and

tan β = \frac{1}{2 m + 1}

then prove that

α + β = \frac{π}{4}

Q. If

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

, show that

α + β = π / 4

Q. If

t a n α = \frac{1}{1 + 2^{- x}}

and

t a n β = \frac{1}{1 + 2^{x + 1}}

, then write the value of

α + β

lying in the interval

(0, \frac{π}{2})

Q. If

tan α = m / (m + 1), tan β = 1 / (2 m + 1)

, show that

α + β = 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 π4Given that, tanα=xx+1 and tanβ=12x+1 Now, tan(α+β)=tanα+tanβ1−tanα.tanβ ⇒tan(α+β)=xx+112x+1 ⇒tan(α+β)=x(2x+1)+x+1(x+1)(2x+1)−x ⇒tan(α+β)=2x2+x+x+12x2+2x+x+1−x ⇒tan(α+β)=2x2+2x+12x2+2x+1 ⇒tan(α+β)=1 ∴α+β=π4

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