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Question
Solve: tan-1 4 x + tan-1 6x `= π/(4)`.
Solution
We have `tan^-1 4 "x" + tan^-1 6"x" = π/(4)`
⇒ `tan ( tan^-1 4"x" + tan^-1 6"x") = tan π/(4)`
⇒ `(tan(tan^-1 4"x") + tan(tan^-1 6"x"))/(1-tan(tan^-1 4"x")·tan(tan^-1 6"x")` = 1
⇒ `(4"x" + 6"x")/(1-4"x" 6"x")` = 1
⇒ `(10"x")/(1 - 24"x"^2)` = 1
⇒ 24x2 + 10x - 1 = 0
⇒ 24x2 + 12x - 2x - 1 = 0
⇒ 12x (2x + 1) - 1 (2x + 1) = 0
⇒ (2x + 1) (12x - 1) = 0
⇒ `"x" = -(1)/(2) , (1)/(12)`
But `"x" = (-1)/(2)` does not satisfy the equation as the LHS will become negative. Therefore, the value of `"x" "is" (1)/(12)`.
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