Question

Find the value of  `tan(sin^-1  3/5 + cot^-1  3/2)`

Answer 1

Let α = `sin^-1 (3/5)`

⇒ sin α = `3/5`

1 – sin²α = `1 - 9/25 = 16/25`

cos²α = `16/25`

⇒ cos α = `4/5`

tan α = `(3/5)/(4/5) = 3/4`

α = `tan^-1 (3/4)`

⇒ `sin^-1 (3/5) = tan^-1(3/4)`

⇒ `tan[sin^-1 (3/5) + cot^-1 (3/2)]`

= `tan[tan^-1  3/4 + tan^-1  2/3]`

= `tan[tan^-1 ((3/4 + 2/3)/(1 - 3/4 xx 2/3))]`

= `tan [tan^-1  (17/12)/(6/12)]`

= `tan[tan^-1 (17/6)]`

= `17/6`