Misc. 8 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Misc 8 Prove that tan-1 1/5 + tan-1 1/7 + tan-1 1/3 + tan-1 1/8 = 𝜋/4 We know that tan-1 x + tan-1 y = tan-1 ((𝐱 + 𝐲 )/(𝟏 − 𝐱𝐲)) Replacing x by 1/5 and y by 1/7 tan-1 1/5 + tan-1 1/7 = tan-1 (1/5 + 1/7)/(1− 1/5 × 1/7) = tan-1 ((7 + 5)/(5(7)))/( (35 − 1)/35 ) = tan-1 (12/35)/( 34/35 ) = tan-1 (6/17) Similarly, Replacing x by 1/3 and y by 1/8 tan-1 1/3 + tan-1 1/8 = tan-1 (1/3 + 1/8)/(1− 1/3 × 1/8) = tan-1 ( (8 + 3)/(3(8)))/( (24 − 1)/24) = tan-1 ( 11/24)/( 23/24) = tan-1 (11/23) Taking L.H.S tan-1 1/5 + tan-1 1/7 + tan-1 1/3 + tan-1 1/8 = (tan-1 1/5 + tan-1 1/7) + ( tan-1 1/3 + tan-1 1/8) = tan-1 (6/17) + tan-1 (11/23) = tan-1 (6/17 + 11/23)/(1 − 6/17 × 11/23) = tan-1 ( (6(23) + 11(17))/(17(23)))/(1 − 66/391) = tan-1 ( (138 + 187)/391)/((391 − 66)/391) = tan-1 ( 325/391)/(325/391) = tan-1 325/391 × (391 )/325 = tan-1 1 = tan-1 1 = tan-1 (tan 𝜋/4) = π/4 = R.H.S Hence proved