Ex 2.2, 4 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Ex 2.2, 4 Prove 2tan-1 1/2 + tan-1 1/7 = tan-1 31/17 First we calculate value of 2tan-1 𝟏/𝟐 We know that 2tan-1x = tan-1 ((𝟐𝐱 )/( 𝟏 − 𝐱^𝟐 )) Replacing x with 1/2 2tan-1 1/2 = tan-1 (2 × 1/2)/(1 − (1/2)2) = tan-1 (1/(1 − 1/4)) = tan-1 (1/((4 − 1)/4)) = tan-1 (1/(3/4)) = tan-1 (4/3) Taking L.H.S. 2tan-1 1/2 + tan-1 1/7 Putting value of 2tan-1 1/2 = tan-1 4/3 + tan-1 1/7 = tan-1 ((4/3 + 1/7 )/( 1− 4/3 × 1/7)) = tan-1 (((4 × 7 +3 × 1 )/( 7 × 3) )/( (7 × 3 − 4)/(7 × 3))) = tan-1 (((28 + 3 )/( 21) )/( ( 21 − 4)/21)) = tan-1 ((31/( 21) )/(17/21)) = tan-1 (31/21 × 21/17)= tan-1 (31/17) = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved