Question

If sin θ = `a/b`, show that `(sectheta + tan theta) = sqrt((b+a)/(b-a))`

Answer 1

Given: `sintheta = a/b`

To show: `sec thetatantheta = sqrt((b + a)/(b - a))`

Using Trigonometric identity,

`costheta = sqrt(1 - sin^2 theta)`

`costheta = sqrt(1 - (a/b)^2)`

`costheta = sqrt((b^2 - a^2)/(b^2))`

`costheta = sqrt((b^2 - a^2)/(b))`

⇒ `sectheta = 1/(costheta) = b/(sqrt(b^2 - a^2))`

⇒ `tan theta = (sintheta)/(costheta) = a/(sqrt(b^2 - a^2))`

Now, 

`sec theta + tan theta`

= `b/(sqrt(b^2 - a^2)) + a/(sqrt(b^2 - a^2))`

= `(b + a)/(sqrt((b - a)(b + a)))`

= `(sqrt(b + a)sqrt(b + a))/(sqrt(b - a)sqrt(b + a))`

= `(sqrt(b + a))/(sqrt(b - a))`

= `sqrt((b + a)/(b - a))`