What is the derivative of y=(1-sec(x))/tan(x)y=1−sec(x)tan(x)?
1 Answer
Answer,
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) , theny'=(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,
y'=(sec(x)-sec^2(x))/(tan^2(x))
Simplifying further, we get
y'=(cos(x)-1)/(sin^2(x))
