If tan−1(3x)+tan−12x=4π, then x is:

tan−1(3x)+tan−1(2x)=4π⇒tan−1{1−3x×2x3x+2x}=4π⇒(1−6x25x)=tan4π⇒1−6x25x=1⇒1−6x2=5x⇒6x2+5x−1=0⇒6x2+6x−x−1=0⇒6x(x+1)−1(x+1)=0⇒(x+1)(6x−1)=0⇒x=−1,x=61Hence, option (a) is correct.

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If ^{-1}(3x)+^{-1} 2x={}{4}, then x is:

Question

If $tan^{- 1} (3 x) + tan^{- 1} 2 x = \frac{π}{4}$ , then $x$ is:

\frac{1}{6}

\frac{1}{3}

\frac{1}{10}

\frac{1}{2}

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Alexander Discussed

tan^{- 1} (3 x) + tan^{- 1} 2 x = \frac{π}{4}

, then

x

is:

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

tan^{- 1} (3 x) + tan^{- 1} (2 x) = \frac{π}{4}

\Rightarrow tan^{- 1} {\frac{3 x + 2 x}{1 - 3 x \times 2 x}} = \frac{π}{4}

\Rightarrow (\frac{5 x}{1 - 6 x ^{2}}) = tan \frac{π}{4}

\Rightarrow \frac{5 x}{1 - 6 x ^{2}} = 1

\Rightarrow 1 - 6 x^{2} = 5 x

\Rightarrow 6 x^{2} + 5 x - 1 = 0

\Rightarrow 6 x^{2} + 6 x - x - 1 = 0

\Rightarrow 6 x (x + 1) - 1 (x + 1) = 0

\Rightarrow (x + 1) (6 x - 1) = 0

\Rightarrow x = - 1, x = \frac{1}{6}

Hence, option

(a)

is correct.

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Question Text	If $tan^{- 1} (3 x) + tan^{- 1} 2 x = \frac{π}{4}$ , then $x$ is:
Answer Type	Text solution:1
Upvotes	150

If tan−1(3x)+tan−12x=4π​, then x is:

Text solutionVerified

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If $tan^{- 1} (3 x) + tan^{- 1} 2 x = \frac{π}{4}$ , then $x$ is: