Question

Show that `"sin"^-1(5/13) + "cos"^-1(3/5) = "tan"^-1(63/16)`

Answer 1

Let `"sin"^-1 (5/13) = "A" => "sin""A" = 5/13`

`therefore "cos""A" = sqrt(1-"sin"^2"A") = sqrt(1-(5/13)^2)`

`sqrt (1-25/169) = sqrt(144/169) = 12/13`

`=> "tan""A" = 5/12`

let `"cos"^-1(3/5) = "B" => "cos""B" = 3/5 `

sin B =`sqrt(1-9/25) = sqrt (16/25) = 4/5`

`therefore "tan""B" = 4/3`

Now , tan (A + B) =` ("tan""A" +"tan""B")/(1 - "tan""A""tan " "B")`

`=> "A" + "B" = "tan"^-1 [[ 5/12 +4/3]/(1-5/12 xx 4/3)]`

`= "tan"^-1 [[(5+16]/12)/( (36 - 20)/36]]`

A +B = `"tan"^-1 [(21/16)/3]`

`=> "sin"^-1 (5/13) + "cos"^-1(3/5) = "tan"^-1(63/16)`