Ex 2.2, 15 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Ex 2.2, 15 If tan-1 (x − 1)/(x − 2) + tan-1 (x + 1)/(x + 2) = 𝜋/4 , then find the value of x. We know that tan-1 x + tan-1 y = tan-1 ((𝐱 + 𝐲 )/( 𝟏 − 𝐱𝐲)) 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 − 2) × (x + 1)/(x + 2))] = tan-1 [(((x − 1) (x + 2) + (x + 1)(x − 2))/((x − 2) (x + 2) ))/(((x − 2) (x + 2) − (x − 1) (x + 1))/((x − 2) (x + 2) ))] = tan-1 [((x − 1) (x + 2) + (x + 1)(x − 2))/((x − 2) (x + 2) ) × ((x − 2) (x + 2))/((x + 2) (x − 2) − (x − 1)(x + 1))] = tan-1 [((x − 1) (x + 2) + (x + 1)(x − 2))/((x + 2) (x − 2) − (x − 1)(x + 1)) ] Using (a + b) (a – b) = a2 – b2 = tan-1 [((x − 1) (x + 2) + (x + 1)(x − 2))/(𝑥2 − 22 −[𝑥2 − 12]) ] = tan-1 [(𝑥 (x + 2) − 1 (x + 2) + x (x − 2) + 1 (x − 2))/(𝑥2 − 4 −𝑥2 + 1) ] = tan-1 [(𝑥2 + 2𝑥 − 𝑥 − 2 + 𝑥2 − 2𝑥 + 𝑥 − 2 )/(𝑥2 − 𝑥2 − 4 + 1) ] Using (a + b) (a – b) = a2 – b2 = tan-1 [((x − 1) (x + 2) + (x + 1)(x − 2))/(𝑥2 − 22 −[𝑥2 − 12]) ] = tan-1 [(𝑥 (x + 2) − 1 (x + 2) + x (x − 2) + 1 (x − 2))/(𝑥2 − 4 −𝑥2 + 1) ] = tan-1 [(𝑥2 + 2𝑥 − 𝑥 − 2 + 𝑥2 − 2𝑥 + 𝑥 − 2 )/(𝑥2 − 𝑥2 − 4 + 1) ] = tan-1 [(𝑥2+ 𝑥2+ 2𝑥− 2𝑥 − 𝑥 + 𝑥 − 2 − 2)/(𝑥2 − 𝑥2 − 4 + 1) ] = tan-1 [(2x2 −4)/(−3)] ∴ tan-1 ((x − 1)/(x − 2)) + tan-1 ((x + 1)/(x + 2)) = tan-1 [(2x2 −4)/(−3)] Given tan-1 (x − 1)/(x − 2) + tan-1 (x + 1)/(x + 2) = 𝜋/4 Putting values tan-1 [(2x2 −4)/(−3)] = π/4 (2x2 −4)/(−3) = tan π/4 (2x2 −4)/(−3) = tan 45° (2x2 −4)/(−3) = 1 2x2 – 4 = − 3 2x2 = − 3 + 4 2x2 = 1 x2 = 1/2 x = ± √(1/2) = ± 1/√2 Thus, x = ± 1/√2