Question

If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.

Answer 1

`sec θ = 41/40`   ...[Given]

∴ `cos θ = 1/(secθ) = 1/(41/40)`

∴ `cos θ = 40/41`

We know that,

sin²θ + cos²θ = 1

∴ `sin^2θ + (40/41)^2 = 1`

∴ `sin^2θ + 1600/1681 = 1`

∴ `sin^2θ = 1 - 1600/1681`

∴ `sin^2θ = (1681- 1600)/1681`

∴ `sin^2θ = 81/1681`

∴ `sin θ = 9/41`   ...[Taking square root of both sides]

Now, cosec θ = `1/(sinθ)`

= `1/((9/41))`

= `41/9`

`cot θ = (cosθ)/(sinθ)`

= `((40/41))/((9/41))`

= `40/9`

∴ `sin θ = 9/41, cot θ = 40/9`, cosec θ = `41/9`