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

1 Answer
Jul 27, 2014

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))#