Question

Show that `2 cot^(-1)  3/2 +  sec^(-1)  13/12 = π/2`

Answer 1

`2 cot^(-1)  3/2 = 2 tan^(-1)  2/3    ...[because cot^(-1) x = tan^(-1) (1/x)]`

`= tan^(-1) [(2 × 2/3)/(1 - (2/3)^2)]  ...[because 2 tan^(-1) x = tan^(-1) ((2x)/(1 - x^2))]`

`= tan^(-1) [(4/3)/(1 - 4/9)]`

`= tan^(-1) (4/3 × 9/5)`

`= tan^(-1)  12/5`    ....(1)

Let `sec^(-1)  13/12  = α`

Then, `sec α = 13/12,  "where"   0 < α < pi/2`

∴ `tan α > 0`

Now, `tan α = sqrt(sec^2 α - 1)`

`tan α = sqrt(169/144 - 1)`

`tan α = sqrt(25/144)`

`tan α = 5/12`

∴ `α = tan^(-1)  5/12 = cot^(-1)  12/5   .....[because tan^(-1) x = cot^(-1) (1/x)]`

∴  `sec^(-1)  13/12 = cot^(-1)  12/5`     .....(2)

Now,

LHS = `2 cot^(-1)  3/2 + sec^(-1)  13/12`

LHS = `tan^(-1)  12/5 + cot^(-1)  12/5`    ...[By (1) and (2)]

LHS = `π/2        .....[because tan^(-1) x + cot^(-1) x - π/2]`

RHS = `π/2`

LHS = RHS

`2 cot^(-1)  3/2 +  sec^(-1)  13/12 = π/2`

Hence proved.