Prove 4tan−1(51)−tan−1(701)+tan−1(991)=4π.

To be proved : 4tan−151−tan−1(701)+tan−1(991)=4πWe know that 2tan−1x=tan−1(1−x22x)So, 2tan−151=tan−1⎝⎛1−(51)22×51⎠⎞=tan−1(125)LHS:4tan−151+tan−1991−tan−1701=(2tan−151−tan−1701)+(2tan−151−tan−1701)+(2tan−151+tan−1991)=(tan−1125−tan−1701)+(tan−1125+tan−1991)=tan−1⎝⎛1+(125)(701)125−701⎠⎞+tan−1⎝⎛1−(125)(991)125+991⎠⎞(∵tan−1x+tan−1y=tan−1(1−xyx+y))=tan−1(52)+tan−1(73)=tan−1⎝⎛1−35652+73⎠⎞=tan−1(1)=4π

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Prove 4^{-1}({1}{5})-^{-1}({1}{70})+^{-1}({1}{99})={}{4}.

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Prove
$4 tan^{- 1} (\frac{1}{5}) - tan^{- 1} (\frac{1}{70}) + tan^{- 1} (\frac{1}{99}) = \frac{π}{4}$ .

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Prove

4 tan^{- 1} (\frac{1}{5}) - tan^{- 1} (\frac{1}{70}) + tan^{- 1} (\frac{1}{99}) = \frac{π}{4}

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To be proved :

4 tan^{- 1} \frac{1}{5} - tan^{- 1} (\frac{1}{70}) + tan^{- 1} (\frac{1}{99}) = \frac{π}{4}

We know that

2 tan^{- 1} x = tan^{- 1} (\frac{2 x}{1 - x ^{2}})

So,

2 tan^{- 1} \frac{1}{5} = tan^{- 1} ⎝ ⎛ \frac{2 \times \frac{1}{5}}{1 - ( \frac{1}{5} ) ^{2}} ⎠ ⎞ = tan^{- 1} (\frac{5}{12})

L H S : 4 tan^{- 1} \frac{1}{5} + tan^{- 1} \frac{1}{99} - tan^{- 1} \frac{1}{70}

= (2 tan^{- 1} \frac{1}{5} - tan^{- 1} \frac{1}{70}) + (2 tan^{- 1} \frac{1}{5} - tan^{- 1} \frac{1}{70}) + (2 tan^{- 1} \frac{1}{5} + tan^{- 1} \frac{1}{99})

= (tan^{- 1} \frac{5}{12} - tan^{- 1} \frac{1}{70}) + (tan^{- 1} \frac{5}{12} + tan^{- 1} \frac{1}{99})

= tan^{- 1} ⎝ ⎛ \frac{\frac{5}{12} - \frac{1}{70}}{1 + ( \frac{5}{12} ) ( \frac{1}{70} )} ⎠ ⎞ + tan^{- 1} ⎝ ⎛ \frac{\frac{5}{12} + \frac{1}{99}}{1 - ( \frac{5}{12} ) ( \frac{1}{99} )} ⎠ ⎞

(∵ tan^{- 1} x + tan^{- 1} y = tan^{- 1} (\frac{x + y}{1 - x y}))

= tan^{- 1} (\frac{2}{5}) + tan^{- 1} (\frac{3}{7})

= tan^{- 1} ⎝ ⎛ \frac{\frac{2}{5} + \frac{3}{7}}{1 - \frac{6}{35}} ⎠ ⎞ = tan^{- 1} (1)

= \frac{π}{4}

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Question Text	Prove $4 tan^{- 1} (\frac{1}{5}) - tan^{- 1} (\frac{1}{70}) + tan^{- 1} (\frac{1}{99}) = \frac{π}{4}$ .
Answer Type	Text solution:1
Upvotes	150

Prove 4tan−1(51​)−tan−1(701​)+tan−1(991​)=4π​.

Text solutionVerified

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Prove
$4 tan^{- 1} (\frac{1}{5}) - tan^{- 1} (\frac{1}{70}) + tan^{- 1} (\frac{1}{99}) = \frac{π}{4}$ .