Question

Prove that cot⁻¹(7) + 2 cot⁻¹(3) = `pi/4`

Answer 1

L.H.S. = cot⁻¹(7) + 2 cot⁻¹(3) 

= cot^–1(7) + cot^–1(3) + cot^–1(3)

= `pi/2 -tan^-1(7) + pi/2 - tan^-1(3) + pi/2 - tan^-1(3)`     .......`[∵ tan^(−1)x + cot^(−1)x = pi/2]`

= `(3pi)/2 - [pi + tan^-1 ((7 + 3)/(1 - 7 xx 3)) + tan^-1(3)]`    .......`[∵ tan^(-1)x + tan^(-1)y = pi + tan^(-1)  (x + y)/(1 - xy), "if"  x, y > 0 and xy > 1]`

= `(3pi)/2 - pi - [tan^-1 (10/-20) + tan^-1(3)]`

= `pi/2 - [tan^-1 (1/2) + tan^-1(3)]`

= `pi/2 - [tan^-1(3) - tan^-1(1/2)]`   .......`[∵ tan^-1(-x) = -tan^-1(x)]`

= `pi/2 -  [tan^-1((3 - 1/2)/(1 + (3)(1/2)))]`

= `pi/2 - [tan^-1((5/2)/(5/2))]`

= `pi/2 - tan^-1(1)`

= `pi/2 - pi/4`

= `pi/4`

= R.H.S.