Question

tan⁻¹x-1/x-2 + tan⁻¹x+1/x+2= π/4,Solve it.

Answer 1

here is ur answer. ..

Answer 2

it is given that, tan⁻¹{(x - 1)/(x - 2)} + tan⁻¹{(x + 1)/(x + 2)} = π/4

we know,

tan⁻¹A + tan⁻¹B = tan⁻¹[(A + B)/(1 - AB)]

where, AB < 1

so, tan⁻¹{(x - 1)/(x - 2)} + tan⁻¹{(x + 1)/(x + 2)} = tan⁻¹[{(x - 1)/(x - 2) + (x + 1)/(x + 2)}/{1 - (x - 1)(x + 1)/(x - 2)(x + 2)}]

= tan⁻¹[{(x - 1)(x + 2) + (x + 1)(x - 2)}/{(x²-2²) - (x² - 1²)}]

= tan⁻¹{x² + x - 2 + x² -x - 2}/(-3)

= tan⁻¹(4 - 2x²)/3

now, tan⁻¹(4 - 2x²)/3 = π/4 ,

or, (4 - 2x²)/3 = tan(π/4)

or, (4 - 2x²)/3 = 1

or, 2x² = 1

or, x = ±1/√2

[ where (x-1)/(x - 2) × (x + 1)/(x + 2) < 1 ⇒(x² - 1)/(x² - 4) < 1

⇒(x² - 1 - x² + 4)/(x² - 4) < 0

⇒3/(x² - 4) < 0

⇒(x² - 4) < 0

or, -2 < x < 2 ]

hence, value of x = 1/√(2) and -1/√(2)