Question

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

Answer 1

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

Explanation:

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)`

Answer 2

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

Explanation:

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)`