FORMULA FOR TAN INVERSE A PLUS TAN INVERSE B
Prove the following results.
Problem 1 :
Solution :
Hence proved.
Problem 2 :
tan-1 17 + 2 tan-1 13 = 𝜋4
Solution :
Given, tan-1 17 + 2 tan-1 13 = 𝜋42 tan-1 x = tan-12x1 - x22 tan-1 13 = tan-12131 - 132= tan-1231 - 19= tan-12389= tan-1 2 3 × 982 tan-1 13 = tan-1 34tan-1 17 + tan-1 34 = 𝜋4 tan-1 x + tan-1 y = tan-1x + y1 - xyx = 17 and y = 34 tan-1 17 + tan-1 34 = tan-117 + 341 - 1734= tan-14 + 21281 - 328= tan-12528 2528= tan-1 2528 × 2825= tan-1(1)tan-1 17 + tan-1 34 = 𝜋4
Hence proved.
Solve the following repressions for x.
Problem 3 :
tan-1 2x + tan-1 3x = n𝜋 + 3𝜋4
Solution :
Given, tan-1 2x + tan-1 3x = n𝜋 + 3𝜋4 tan-1 x + tan-1 y = tan-1x + y1 - xyx = 2x and y = 3x tan-1 2x + tan-1 3x = tan-12x + 3x1 - (2x)(3x)⟹tan-15x1 - 6x2 = n𝜋 + 3𝜋4⟹5x1 - 6x2 = tann𝜋 + 3𝜋4⟹5x1 - 6x2 = -15x = -11 - 6x25x = -1 + 6x26x2 - 5x - 1 = 06x2 - 6x + 1x - 1= 0 6x(x - 1) + 1(x - 1) = 0(6x + 1) (x - 1) = 06x + 1 = 0 and x - 1 = 06x = -1 and x = 1x = -16x = 1 is not satisfying the equation.
Problem 4 :
tan-1 (x + 1) + tan-1 (x - 1) =tan-1831
Solution :
Given, tan-1 (x + 1) + tan-1 (x - 1) =tan-1831 tan-1 x + tan-1 y = tan-1x + y1 - xyx = x + 1 and y = x - 1 tan-1 (x + 1) + tan-1 (x - 1) = tan-1x + 1 + x - 11 - (x + 1)(x - 1)⟹ tan-1x + 1 + x - 11 - (x + 1)(x - 1) = tan-1831⟹2x1 - x2 - 1 = 831⟹2x2 - x2 = 83162x = 16 - 8x28x2 + 62x - 16 = 024x2 + 31x - 8 = 04x2 + 31x - 8 = 0 4x2 + 32x - x - 8 = 0 4x(x + 8) - 1(x + 8) = 0(4x - 1) (x + 8) = 04x - 1 = 0 and x + 8 = 04x = 1 and x = -8x = 14x = -8 is not satisfying the equation.