Formulae based

Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

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Misc 14 Solve tan-1 (1 β x)/(1 + x) = 1/2 tan-1 x, (x > 0) tan-1 (1 β x)/(1 + x) = 1/2 tan-1 x 2 tan-1 ((1 β x)/(1 + x)) = tan-1 x tan-1 [(2 ((1 β π₯)/(1 + π₯)))/(1 β ((1 β π₯)/(1 + π₯ ))^2 )] = tan-1 x We know that 2 tan-1 x = tan-1 ((ππ )/(π β π±^π )) Replacing x by (1 β π₯)/(1 + π₯) tan-1 [((2 (1 β π₯))/((1 + π₯)))/(((1 + π₯)2 β ( 1 βπ₯ )2)/(1 + π₯ )^2 )] = tan-1 x tan-1 [(2 (1 β π₯))/((1 + π₯)) Γ ((1 + π₯))/((1 + π₯)2 β (1 β π₯)2)] = tan-1 x tan-1 [(2 (1 β π₯) (1 + π₯))/((1 + π₯)2 β (1 β π₯)2)] = tan-1 x Using (a + b) (a β b) = a2 β b2 tan-1 [ (2 (1 β π₯2) )/((1 + π₯ + 1 β π₯) (1+ π₯ β 1 + π₯) )] = tan-1 x tan-1 [ (2 (1 β π₯2) )/((1 +1 β π₯ β π₯) (π₯ + π₯ β 1 + 1) )] = tan-1 x tan-1 [(2 (1 β π₯2))/(4 (1) (π₯) )] = tan-1 x tan-1 [(1 β π₯2)/2π₯] = tan-1 x Comparing values (1 β π₯2)/2π₯ = x 1 β x2 = 2x2 1 β x2 β 2x2 = 0 1 β 3x2 = 0 3x2 = 1 x2 = 1/3 x = Β± 1/β3 x = (β 1 )/β3 is not possible because it is Given that x > 0 Hence, x = ( π )/βπ