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Prove that 2^{-1} {1}{2} + ^{-1} {1}{7} = ^{-1} {31}{17

Question

Prove that

$2 tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{7} = tan^{- 1} \frac{31}{17}$

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Prove that

2 tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{7} = tan^{- 1} \frac{31}{17}

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Text solutionVerified

Consider L.H.S

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

We know that,

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

= tan^{- 1} ⎝ ⎛ \frac{2 \times \frac{1}{2}}{1 - ( \frac{1}{2} ) ) ^{2}} ⎠ ⎞ + tan^{- 1} (\frac{1}{7})

= tan^{- 1} ⎝ ⎛ \frac{1}{1 - \frac{1}{4}} ⎠ ⎞ + tan^{- 1} (\frac{1}{7})

= tan^{- 1} ⎝ ⎛ \frac{1}{\frac{4 - 1}{4}} ⎠ ⎞ + tan^{- 1} (\frac{1}{7})

= tan^{- 1} ⎝ ⎛ \frac{1}{\frac{3}{4}} ⎠ ⎞ + tan^{- 1} (\frac{1}{7})

= tan^{- 1} (\frac{4}{3}) + tan^{- 1} (\frac{1}{7})

We know that

tan^{- 1} A + tan^{- 1} B = tan^{- 1} (\frac{A + B}{1 - A B})

= tan^{- 1} ⎝ ⎛ \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}} ⎠ ⎞

= tan^{- 1} ⎝ ⎛ \frac{\frac{28 + 3}{21}}{1 - \frac{4}{21}} ⎠ ⎞

= tan^{- 1} ⎝ ⎛ \frac{\frac{31}{21}}{\frac{21 - 4}{21}} ⎠ ⎞

= tan^{- 1} (\frac{31}{17}) =

R.H.S

Hence proved.

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Question Text	Prove that $2 tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{7} = tan^{- 1} \frac{31}{17}$
Answer Type	Text solution:1
Upvotes	150