Question

Answer the following:

If sinθ = `(x^2 - y^2)/(x^2 + y^2)` then find the values of cosθ, tanθ in terms of x and y.

Answer 1

Given, sinθ = `(x^2 - y^2)/(x^2 + y^2)`

We know that,

cos²θ = 1 – sin²θ

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

= `((x^2 + y^2)^2 - (x^2 - y^2)^2)/((x^2 + y^2)^2`

= `(x^4+2x^2y^2+y^4-(x^4-2x^2y^2+y^4))/((x^2+y^2)^2)`

∴ cos²θ = `(4x^2y^2)/((x^2 + y^2)^2`

∴ cosθ = `±(2xy)/((x^2 + y^2))`

tanθ = `sintheta/costheta`

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

= `±(x^2 - y^2)/(2xy)`

Hence, cosθ = `±(2xy)/(x^2 + y^2)` and tanθ = `±(x^2 - y^2)/(2xy)`