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प्रश्न
Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`
उत्तर
`cot^-1 x - cot^-1 (x + 2) = pi/12`
`tan^-1[1/x] - tan^-1 [1/(x + 2)] = pi/12`
⇒ `tan^-1 [(1/x - 1/(x + 2))/(1 + (1/x)(1/(x + 2)))] = pi/12`
⇒ `(x + 2 - x)/(x(x + 2) + 1) = tan pi/12`
⇒ `2/(x^2 + 2x + 1) = tan15^circ`
We know that, tan 15° = `2 - sqrt(3)`
⇒ `2/(x^2 + 2x + 1) = 2 - sqrt(3)`
⇒ `x^2 + 2x + 1 = 2/(2 - sqrt(3)`
⇒ `(x + 1)^2 = 2/(2 - sqrt(3)) xx [(2 + sqrt(3))/(2 + sqrt(3))]`
⇒ `(x + 1)^2 = (2(2 + sqrt(3)))/(4 - 3)`
⇒ (x + 1)2 = `4 + 2sqrt(3)`
⇒ (x + 1)2 = `1 + 3 + 2sqrt(3)`
⇒ (x + 1)2 = `(1 + sqrt(3))^2`
⇒ x + 1 = `1 + sqrt(3)`
∴ x = `sqrt(3)`
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