prove that 2 tan-1 (root of a+b/a-b tn theta/2) = cos -1 (acos theta b/a b cos theta) - Maths - Inverse Trigonometric Functions

Question

prove that 2 tan-1 (root of a+b/a-b tn theta/2) = cos -1 (acos
theta b/a b cos theta)

Vipin Verma · Accepted Answer

2tan-1(a-ba+btanθ2) = cos-1acosθ +ba+bcosθAs 2tan-1x
=cos-11-x21+x2 (1)Let x =a-ba+btanθ2 , so put this in (1)So
RHS = cos-11-(a-ba+btanθ2)21+(a-ba+btanθ2)2
=cos-11-a-ba+btan2θ21+a-ba+btan2θ2= cos-1a[1-tan2θ2]
+b[1+tan2θ2] a[1+tan2θ2] +b[1-tan2θ2]
=cos-1a[cos2θ2-sin2θ2]
+b[cos2θ2+sin2θ2]a[cos2θ2+sin2θ2]
+b[cos2θ2-sin2θ2] =cos-1acosθ +bbcosθ +aHence
proved