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Question
Solve the following equation for x:
tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
Solution
Given: tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
Take LHS
tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
We know that, Formula
tan−1 x + tan-1 y = tan-1 `(x + y)/(1 - xy)`
Thus,
`=> tan^-1 ((x + 1)+(x - 1))/(1 -(x + 1)xx(x - 1)) = tan^-1 8/31`
`=> tan^-1 (2x)/(1-(x^2 - 1)) = tan^-1 8/31`
`=> tan^-1 (2x)/(1 - x^2 + 1) = tan^-1 8/31`
`=> (2x)/(1 - x^2 + 1) = 8/31`
⇒ 62x = 8 − 8x2 + 8
⇒ 4x2 + 62x − 16 = 0
⇒ 6x2 + 31x − 8 = 0
⇒ 4x(x + 8) − 1(x + 8) = 0
⇒ (4x − 1)(x + 8) = 0
⇒ 6x + 1 = 0 or x − 1 = 0
⇒ x = `1/4` or x = −8
Since,
x = `1/4` ∈ `(-sqrt2, sqrt2)`
So,
x = `1/4` is the root of the given equation
Therefore,
x = `1/4`
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