tan−1(41)+2tan−1(51)+tan−1(61)+tan−1(x1)=4π.

tan⎝⎛1−x1×41x1+41⎠⎞+tan−1⎩⎨⎧1−(51)22×51⎭⎬⎫+tan−161=4π⇒tan−1(4x−14+x)+tan−1(125)+tan−161=4π⇒tan−1(4x−14+x)+tan−1⎩⎨⎧1−125×61126+61⎭⎬⎫=4π⇒tan−1(4x−14+x)+tan−1(72−530+12)=4π⇒tan−1(4x−14+x)+tan−1(6742)=4π⇒tan−1⎩⎨⎧1−(4x−14+x)(6742)4x−14+x+6742⎭⎬⎫=4π⇒(4x−1)67−42(4+x)67(4+x)+42(4x−1)=tan4π⇒268x−67−168−42x268+67x+167x−42=1⇒267+67x+168x−42=268x−67−168−42x⇒9x=−461⇒x=9−461

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^{-1}({1}{4})+2^{-1}({1}{5})+^{-1}({1}{6})+^{-1}({1}{x})={}{4}

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

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

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tan ⎝ ⎛ \frac{\frac{1}{x} + \frac{1}{4}}{1 - \frac{1}{x} \times \frac{1}{4}} ⎠ ⎞ + tan^{- 1} ⎩ ⎨ ⎧ \frac{2 \times \frac{1}{5}}{1 - ( \frac{1}{5} ) ^{2}} ⎭ ⎬ ⎫ + tan^{- 1} \frac{1}{6} = \frac{π}{4}

\Rightarrow tan^{- 1} (\frac{4 + x}{4 x - 1}) + tan^{- 1} (\frac{5}{12}) + tan^{- 1} \frac{1}{6} = \frac{π}{4}

\Rightarrow tan^{- 1} (\frac{4 + x}{4 x - 1}) + tan^{- 1} ⎩ ⎨ ⎧ \frac{\frac{6}{12} + \frac{1}{6}}{1 - \frac{5}{12} \times \frac{1}{6}} ⎭ ⎬ ⎫ = \frac{π}{4}

\Rightarrow tan^{- 1} (\frac{4 + x}{4 x - 1}) + tan^{- 1} (\frac{30 + 12}{72 - 5}) = \frac{π}{4}

\Rightarrow tan^{- 1} (\frac{4 + x}{4 x - 1}) + tan^{- 1} (\frac{42}{67}) = \frac{π}{4}

\Rightarrow tan^{- 1} ⎩ ⎨ ⎧ \frac{\frac{4 + x}{4 x - 1} + \frac{42}{67}}{1 - ( \frac{4 + x}{4 x - 1} ) ( \frac{42}{67} )} ⎭ ⎬ ⎫ = \frac{π}{4}

\Rightarrow \frac{67 ( 4 + x ) + 42 ( 4 x - 1 )}{( 4 x - 1 ) 67 - 42 ( 4 + x )} = tan \frac{π}{4}

\Rightarrow \frac{268 + 67 x + 167 x - 42}{268 x - 67 - 168 - 42 x} = 1

\Rightarrow 267 + 67 x + 168 x - 42 = 268 x - 67 - 168 - 42 x

\Rightarrow 9 x = - 461

\Rightarrow x = \frac{- 461}{9}

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Question Text	$tan^{- 1} (\frac{1}{4}) + 2 tan^{- 1} (\frac{1}{5}) + tan^{- 1} (\frac{1}{6}) + tan^{- 1} (\frac{1}{x}) = \frac{π}{4}$ .
Answer Type	Text solution:1
Upvotes	150

tan−1(41​)+2tan−1(51​)+tan−1(61​)+tan−1(x1​)=4π​.

Text solutionVerified

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