Prove that:4tan−151−tan−1701+tan−1991=4π

Question

Prove that:4tan−151​−tan−1701​+tan−1991​=4π​

Filo · Accepted Answer

Let tan−151​=θ⟹tanθ=51​As we know thattan2A=1−tan2A2tanA​So tan2θ=1−tan2A2tanA​=1−(51​)22×51​​=1−251​52​​=125​⟹tan4θ=1−tan22θ2tan2θ​=1−(125​)22×125​​=119120​⟹4θ=tan−1(119120​)⟹4 tan−151​=tan−1119120​4tan−151​−tan−1701​+tan−1991​=tan−1119120​−tan−1701​+tan−1991​                                               =tan−11+119120​×701​119120​−701​​+tan−1991​  (∵tan−1a−tan−1b=tan−11+aba−b​,ab>−1)                                             =tan−15049​+tan−1991​                                             =tan−11−5049​×991​5049​+991​​                            (∵tan−1a+tan−1b=tan−11−aba+b​,ab<1)                                            =tan−11=4π​∴  4tan−151​−tan−1701​+tan−1991​=4π​ Hence Proved

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

Prove that:
$4 tan^{- 1} \frac{1}{5} - tan^{- 1} \frac{1}{70} + tan^{- 1} \frac{1}{99} = \frac{π}{4}$

Text solutionVerified

Practice questions from similar books

Prove that:4tan−151​−tan−1701​+tan−1991​=4π​

Text solutionVerified

Practice questions from similar books

Prove that:
$4 tan^{- 1} \frac{1}{5} - tan^{- 1} \frac{1}{70} + tan^{- 1} \frac{1}{99} = \frac{π}{4}$