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

Jul 27, 2014

Answer, $y ' = \frac{\cos \left(x\right) - 1}{{\sin}^{2} \left(x\right)}$

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

Explanation (I), Simplifying the expression

$y = \frac{1 - \sec \left(x\right)}{\tan} \left(x\right) = \frac{1}{\tan} \left(x\right) - \sec \frac{x}{\tan} \left(x\right)$

$y = \cot \left(x\right) - \frac{1}{\cos} \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right) = \cot \left(x\right) - \csc \left(x\right)$

$y ' = \left(\cot \left(x\right) - \csc \left(x\right)\right) '$

$y ' = - {\csc}^{2} \left(x\right) + \csc \left(x\right) \cot \left(x\right)$

$y ' = \csc \left(x\right) \left(\cot \left(x\right) - \csc \left(x\right)\right)$

$y ' = \frac{1}{\sin} \left(x\right) \left(\cos \frac{x}{\sin} \left(x\right) - \frac{1}{\sin} \left(x\right)\right)$

$y ' = \frac{\cos \left(x\right) - 1}{{\sin}^{2} \left(x\right)}$

Explanation (II)

This can also be solved using Quotient Rule

Which is ,

$y = f \frac{x}{g} \left(x\right)$, then $y ' = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

In same way, for the problem,

$y ' = \frac{\left(1 - \sec \left(x\right)\right) ' \tan \left(x\right) - \left(1 - \sec \left(x\right)\right) \left(\tan \left(x\right)\right) '}{{\tan}^{2} \left(x\right)}$

$y ' = \frac{\left(- \sec \left(x\right) \tan \left(x\right)\right) \tan \left(x\right) - \left(1 - \sec \left(x\right)\right) {\sec}^{2} \left(x\right)}{{\tan}^{2} \left(x\right)}$

$y ' = \frac{- \sec \left(x\right) {\tan}^{2} \left(x\right) - {\sec}^{2} \left(x\right) + {\sec}^{3} \left(x\right)}{{\tan}^{2} \left(x\right)}$

If we club first and last term,

$y ' = \frac{\sec \left(x\right) \left({\sec}^{2} \left(x\right) - {\tan}^{2} \left(x\right)\right) - {\sec}^{2} \left(x\right)}{{\tan}^{2} \left(x\right)}$

Using Trigonometric Identity, ${\sec}^{2} x - {\tan}^{2} x = 1$

$y ' = \frac{\sec \left(x\right) - {\sec}^{2} \left(x\right)}{{\tan}^{2} \left(x\right)}$

Simplifying further, we get

$y ' = \frac{\cos \left(x\right) - 1}{{\sin}^{2} \left(x\right)}$