Solve - tan-14x-tan-16x-4

Tan^ 14x+tan^ 16x=π4 ⇒tan^ 1(4x+6x1 24x^2)=π4, provided 4x× 6x lt;1 ⇒4x+6x1 24x^2=tan^ 1(π4) ⇒10x1 24x^2=1 ⇒ (12x 1)(2x+1)=0 ⇒ x=112, 12 For x=112, 4x ...

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Standard XII

Mathematics

Homogeneous Differential Equation

Solve : tan-1...

Question

$Solve : {tan}^{- 1} 4 x + {tan}^{- 1} 6 x = \frac{π}{4}$

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Solution

${tan}^{- 1} 4 x + {tan}^{- 1} 6 x = \frac{π}{4}$
$\Rightarrow {tan}^{- 1} (\frac{4 x + 6 x}{1 - 24 x^{2}}) = \frac{π}{4}, provided 4 x \times 6 x < 1 \Rightarrow \frac{4 x + 6 x}{1 - 24 x^{2}} = {tan}^{- 1} (\frac{π}{4}) \Rightarrow \frac{10 x}{1 - 24 x^{2}} = 1 \Rightarrow (12 x - 1) (2 x + 1) = 0 \Rightarrow x = \frac{1}{12}, \frac{- 1}{2}$

$For x = \frac{1}{12}, 4 x \times 6 x = \frac{1}{6} < 1$
$For x = \frac{- 1}{2}, 4 x \times 6 x = 6 ≮ 1$
Therefore, the solution of the given equation is $x = \frac{1}{12}$

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Solve : {tan}^{- 1} 4 x + {tan}^{- 1} 6 x = \frac{π}{4}

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Q. Prove that:

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

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{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

Q. The expression

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}})

in the interval

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simlifies to:

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Solve :tan−14x+tan−16x=π4

tan−14x+tan−16x=π4 ⇒tan−1(4x+6x1−24x2)=π4, provided 4x×6x<1⇒4x+6x1−24x2=tan−1(π4)⇒10x1−24x2=1⇒(12x−1)(2x+1)=0⇒x=112,−12 For x=112, 4x×6x=16<1 For x=−12, 4x×6x=6≮1 Therefore, the solution of the given equation is x=112

$Solve : {tan}^{- 1} 4 x + {tan}^{- 1} 6 x = \frac{π}{4}$