tan−1(21tan2A ) +tan−1(cotA ) +tan−1(cot3A ) ={0,if4π <A<2π π,if0<A<2π

tan−1(21tan2A)+tan−1(1−cot4AcotA+cot3A)tan−1(2(1−tan2x)2tanA)+tan1(1−cot2cotA)⇒tan−1(1−tan2AtanA)+tan−1⎝⎛tan2Atan2A−1tanA1⎠⎞⇒tan−1(1−tan2AtanA)+tan−1(tan2A−1tanA)⇒ if 4π<A<2π⇒tan−1(1−tan2AtanA)−tan−1(1−tan2AtanA)=0if 0<A<2π⇒tan−1(1−tan2AtanA)+cot−1(1−tan2AtanA)=π

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^{ -1 }{ ( { 1 }{ 2 } { 2A } ) } + ^{ -1 }{ ( { A } )

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$tan^{- 1} (\frac{1}{2} tan 2 A) + tan^{- 1} (cot A) + tan^{- 1} (cot^{3} A) = {0, i f \frac{π}{4} < A < \frac{π}{2} π, i f 0 < A < \frac{π}{2}$

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tan^{- 1} (\frac{1}{2} tan 2 A) + tan^{- 1} (cot A) + tan^{- 1} (cot^{3} A) = {0, i f \frac{π}{4} < A < \frac{π}{2} π, i f 0 < A < \frac{π}{2}

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Text solutionVerified

tan^{- 1} (\frac{1}{2} tan 2 A) + tan^{- 1} (\frac{cot A + cot 3 A}{1 - cot 4 A})

tan^{- 1} (\frac{2 tan A}{2 ( 1 - tan ^{2} x )}) + tan^{1} (\frac{cot A}{1 - cot ^{2}})

\Rightarrow tan^{- 1} (\frac{tan A}{1 - tan ^{2} A}) + tan^{- 1} ⎝ ⎛ \frac{\frac{1}{tan A}}{\frac{tan ^{2} A - 1}{tan ^{2} A}} ⎠ ⎞

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

\Rightarrow

\frac{π}{4} < A < \frac{π}{2}

\Rightarrow tan^{- 1} (\frac{tan A}{1 - tan ^{2} A}) - tan^{- 1} (\frac{tan A}{1 - tan ^{2} A}) = 0

0 < A < \frac{π}{2}

\Rightarrow tan^{- 1} (\frac{tan A}{1 - tan ^{2} A}) + cot^{- 1} (\frac{tan A}{1 - tan ^{2} A})

= π

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Question Text	$tan^{- 1} (\frac{1}{2} tan 2 A) + tan^{- 1} (cot A) + tan^{- 1} (cot^{3} A) = {0, i f \frac{π}{4} < A < \frac{π}{2} π, i f 0 < A < \frac{π}{2}$
Answer Type	Text solution:1
Upvotes	156

tan−1(21​tan2A ) +tan−1(cotA ) +tan−1(cot3A ) ={0,if4π ​<A<2π ​ π,if0<A<2π ​ ​

Text solutionVerified

Practice questions from similar books

$tan^{- 1} (\frac{1}{2} tan 2 A) + tan^{- 1} (cot A) + tan^{- 1} (cot^{3} A) = {0, i f \frac{π}{4} < A < \frac{π}{2} π, i f 0 < A < \frac{π}{2}$