The number of real solutions of the equation tan ( ∑ r = 1 5 cot − 1 ( 2 r 2 ) ) = 6 x + 5 5 x + 6 , is

cot − 1 ( 2 r 2 ) = tan − 1 { 1 + ( 2 r + 1 ) ( 2 r − 1 ) ( 2 r + 1 ) − ( 2 r − 1 ) } ⇒ 6 5 = 6 x + 5 5 x + 6 No real solution = 0 .

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MathematicsInverse Trigonometric FunctionsSummation of seriesMedium2 minQB

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The number of real solutions of the equation $tan (\sum_{r = 1}^{5} cot^{- 1} (2 r^{2})) = \frac{5 x + 6}{6 x + 5}$ , is

Answer:

Solution:

$cot^{- 1} (2 r^{2}) = tan^{- 1} {\frac{( 2 r + 1 ) - ( 2 r - 1 )}{1 + ( 2 r + 1 ) ( 2 r - 1 )}} \Rightarrow \frac{5}{6} = \frac{5 x + 6}{6 x + 5}$

No real solution = 0 .

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:Summation of series

⏱ 2mℹ️ Source: QB

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