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

Question

Christopher Allers · Accepted Answer

#d/dx(1/(secx-tanx))=1/(1-sinx)#

Note that:

#1/(secx-tanx)=1/(1/cosx-sinx/cosx)=cosx/(1-sinx)#

To differentiate this, we will use the quotient rule, which states that when we have the functions #f# and #g# divided by one another:

#d/dx(f/g)=(f^'*g-f*g^')/g^2#

So here, we see that #f=cosx# so #f^'=-sinx#, and #g=1-sinx# so #g^'=-cosx#. Thus:

#d/dx(cosx/(1-sinx))=((-sinx)(1-sinx)-(cosx)(-cosx))/(1-sinx)^2#

Simplifying:

#=(-sinx+sin^2x+cos^2x)/(1-sinx)^2#

Since #sin^2x+cos^2x=1#:

#=(1-sinx)/(1-sinx)^2#

#=1/(1-sinx)#

Charles Anctil · Answer

Mason has given a fine answer, here's a bit of a tricky one.

One of the Pythagorean trigonometric identities is #tan^2 x + 1 = sec^2 x# Which also gets us #sec^2 x - tan^2 x = 1#. And factoring the difference of squares on the left: #(secx-tanx)(sec x +tan x)# (The is the same bit of algebra we use to rationalize fractions involving #sqrta - b#. In this case applied to trigonometry.) #1/((sec x - tan x )) * ((sec x + tan x))/((secx + tan x)) = secx + tanx#. Now use the differentiation rulles for these trig functions to get #d/dx(1/(sec x - tan x)) = secx tan x + sec^2 x# Comparing answers: #secx tan x + sec^2 x = 1/cos x sinx/cosx + 1/cos^2 x# # = (sinx + 1)/cos^2 x# # = (1+sinx)/(1-sin^2 x)# # = 1/(1-sinx)#

Cameron Alexander · Answer

undefined To find the derivative of $ \frac{1}{\sec(x) - \tan(x)} $, apply the quotient rule.

Let $ u(x) = 1 $ and $ v(x) = \sec(x) - \tan(x) $.

Using the quotient rule:
$$ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Now, find the derivatives of $ u(x) $ and $ v(x) $:
$$ u'(x) = 0 $$
$$ v'(x) = \sec(x)\tan(x) - \sec^2(x) $$

Plug these derivatives into the quotient rule:
$$ \frac{d}{dx}\left(\frac{1}{\sec(x) - \tan(x)}\right) = \frac{0 \cdot (\sec(x) - \tan(x)) - 1 \cdot (\sec(x)\tan(x) - \sec^2(x))}{(\sec(x) - \tan(x))^2} $$

Simplify:
$$ \frac{-\sec(x)\tan(x) + \sec^2(x)}{(\sec(x) - \tan(x))^2} $$

This is the derivative of $ \frac{1}{\sec(x) - \tan(x)} $.