Question

if sec^-1  x = cosec^-1  v. show that `1/x^2 + 1/y^2 = 1`

Answer 1

 sec-1 x=cos ec-1 y; x ∈ R-(-1,1) and y ∈ R - (-1,1)
`cos^-1 (1/x)= sin^-1(1/y)`

`cos^-1 (1/x)=pi/2-cos^-1(1/y)`

`cos-1(1/x)+cos^-1 (1/y)=pi/2`

`therefore cos ^-1 (1/x xx1/y-sqrt(1-1/x^2) xx sqrt(1-1/y^2))=pi/2`

`1/x xx 1/y - sqrt((1- 1/x^2)(1-1/y^2)) = cos  pi/2`

`1/(xy) - sqrt(1-1/y^2 - 1/x^2 + 1/(x^2 y^2) = 0`

`1/(xy)= sqrt(1-1/y^2 - 1/x^2+ 1/(x^2y^2)`

Squaring both sides
`1/(x^2y^2) =  1- 1/y^2 - 1/x^2 + 1/x^2y^2`

`therefore 1/x^2 + 1/y^2 = 1`