Question

What is the derivative of `y=(1-sec(x))/tan(x)`?

Answer 1

Answer, `y'=(cos(x)-1)/(sin^2(x))`

This problem can be solved by two methods, as mentioned below :

Explanation (I), Simplifying the expression

`y=(1-sec(x))/tan(x) = 1/tan(x)-sec(x)/tan(x)`

`y=cot(x)-1/cos(x)*cos(x)/sin(x)=cot(x)-csc(x)`

`y'=(cot(x)-csc(x))'`

`y'=-csc^2(x)+csc(x)cot(x)`

`y'=csc(x)(cot(x)-csc(x))`

`y'=1/sin(x)(cos(x)/sin(x)-1/sin(x))`

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

Explanation (II)

This can also be solved using Quotient Rule

Which is ,

`y=f(x)/g(x)`, then `y'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2`

In same way, for the problem,

`y'=((1-sec(x))'tan(x)-(1-sec(x))(tan(x))')/(tan^2(x))`

`y'=((-sec(x)tan(x))tan(x)-(1-sec(x))sec^2(x))/(tan^2(x))`

`y'=(-sec(x)tan^2(x)-sec^2(x)+sec^3(x))/(tan^2(x))`

If we club first and last term,

`y'=(sec(x)(sec^2(x)-tan^2(x))-sec^2(x))/(tan^2(x))`

Using Trigonometric Identity, `sec^2x-tan^2x=1`

`y'=(sec(x)-sec^2(x))/(tan^2(x))`

Simplifying further, we get

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