# Proving a trigonometric identity: $\frac{\cos x}{1-\sin x} -\tan x = \sec x$

I am trying to prove a trig identity that is confusing me. The identity is $$\frac{\cos(x)}{(1-\sin(x))}-\tan(x)=\sec(x)$$ Here is my attempt.

I did $$\frac{\cos(x)}{(1-\sin^2(x))}=\frac{\cos(x)}{\cos^2(x)}=\frac{1}{\cos(x)}=\sec(x)=(\sec(x)+\tan(x))(1+\sin(x))\\\sec(x)=\sec(x)+\sec(x)\sin(x)+\tan(x)+\tan(x)\sin(x)\\0=\tan(x)+\tan(x)+\tan(x)\sin(x)$$ but this does not make sense to me. Can somebody please help me with this thing?

• @YagnaPatel traditionally when verifying trig identities we only manipulate one side of the equation. May 14, 2015 at 1:12
• You misread the problem. It is $\frac{\cos x}{1-\sin x}$, so your solution started with the wrong expression. May 14, 2015 at 3:02
• so many upvotes? What for?
– MonK
May 14, 2015 at 10:21
• @ben I wouldn't say that's really the case. You can manipulate just one side of the equations, but it's absolutely valid to work with both. This is particularly useful if you aren't all that proficient. Working from both sides let you use each and every identity you can think of and quite often you get that little crucial step in the middle because your two sides have become close enough to remind you of something.
– DRF
May 14, 2015 at 21:05
• @User58220: Why on earth not? The only issue to watch out for with cross-multiplication is that it can introduce extraneous solutions where either (or both) of the denominators is zero. But that has nothing specifically to do with trig. May 14, 2015 at 23:08

For fun, I found a "trigonograph" of this identity (for acute $\theta$).

In the diagram, $\overline{AB}$ is tangent to the unit circle at $P$. The "trig lengths" (except for $|\overline{AQ}|$) should be clear. We note that $\angle BPR \cong \angle RPP^\prime$, since these inscribed angles subtend congruent arcs $\stackrel{\frown}{PR}$ and $\stackrel{\frown}{RP^\prime}$. Very little angle chasing gives that $\triangle APQ$ is isosceles, with $\overline{AP} \cong \overline{AQ}$ (justifying that last trig length). Then, $$\triangle SPR \sim \triangle OQR \implies \frac{|\overline{SP}|}{|\overline{SR}|} = \frac{|\overline{OQ}|}{|\overline{OR}|} \implies \frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta+\tan\theta}{1}$$

We have
\begin{align}\frac{\cos x}{1-\sin x}-\tan x & = \frac{\cos x}{1-\sin x}-\frac{\sin x}{\cos x} \\[1.5ex] & =\frac{\cos^2 x-\sin x(1-\sin x)}{\cos x(1-\sin x)} \\[1.5ex] & =\frac{\cos^2 x+\sin^2 x-\sin x}{\cos x(1-\sin x)} \\[1.5ex] & =\frac{1-\sin x}{\cos x(1-\sin x)} \\[1.5ex] & =\frac{1}{\cos x} \\[2.8ex] & =\sec x\end{align}

Hint

$$\frac{\cos(x)}{1-\sin(x)}-\tan(x)=\frac{\cos(x)}{1-\sin(x)}-\frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)-\sin(x)+\sin^2(x)}{\cos(x)(1-\sin(x))}$$

Now what is $\cos^2(x)+\sin^2(x)$?

• Thank you for the hint. This really help me. May 14, 2015 at 1:36

$$1 = \sec^2 x - \tan^2 x = (\sec x + \tan x )(\sec x - \tan x)$$ dividing by the second factor on the RHS: $$\frac1{\sec x - \tan x} = \sec x + \tan x$$ multiplying LHS numerator and denominator by $\cos x$ and bringing $\tan x$ over to the LHS from RHS:

$$\frac{\cos x}{1-\sin x} - \tan x = \sec x$$

I am going to abuse the equality sign a little, and manipulate both sides at once. I find that a bit more intuitive

Required to Prove:

$$\dfrac{\cos(x)}{1-\sin(x)}-\tan(x)=\sec(x)$$

Move to "usual" trig functions

$$\dfrac{\cos(x)}{1-\sin(x)}-\dfrac{\sin(x)}{\cos(x)}=\dfrac{1}{\cos(x)}$$

There are $\cos$'s on the bottom line, so lets simplify by multiplying both sides by $\cos(x)$

$$\dfrac{\cos^{2}(x)}{1-\sin(x)}-\sin(x)=1$$

Move the lonely $\sin(x)$

$$\dfrac{\cos^{2}(x)}{1-\sin(x)}=1+\sin(x)$$

The denominator on the bottom of left hand side would be good to have $\cos^2(x)=1-\sin^2(x)=(1-\sin(x))(1+\sin(x))$. and we can always multiple any term by $1=\dfrac{1+\sin(x)}{1+\sin(x)}$

$$\dfrac{\cos^{2}(x)}{1-\sin(x)}\dfrac{1+\sin(x)}{1+\sin(x)}=1+\sin(x)$$

$$\dfrac{\cos^{2}(x)(1+\sin(x))}{1-\sin^{2}(x)}=1+\sin(x)$$

$$\dfrac{\cos^{2}(x)(1+\sin(x))}{\cos^{2}(x)}=1+\sin(x)$$

$$1+\sin(x)=1+\sin(x)$$

• I don't understand this reasoning. You start with something that you don't know is true and from this basis deduce something that is obviously true, but how does this prove that the original thing is true? E.g. I could start with 1 = 3, multiply both sides by 0 and get 0 = 0. This doesn't mean that 1 = 3 is true. There's obviously something I'm missing in your logic. May 15, 2015 at 6:53
• @CharlesBailey You are correct. This is not a proof, (as I said I abuse the Equals symbol), it is more a wave in the direction of a proof. I believe it can be shown that I never end up multiplying either side by something that is zero (etc), except with the original statement is already not defined (as x=pi/4+k*pi/2), and that thus it does show the identity is correct.) However this is not a proper form for a proof at all, because it is not Obvious that I never break a rule like multiplying each side by zero. May 15, 2015 at 7:33
• @CharlesBailey There is nothing at all wrong with this answer, provided it is read backwards and one assumes that everything which appears in a denominator and in a "multiply by $x$" step is nonzero. Because then you are starting with something that is true and deducing something that we want to show. (People don't seem to realize that this is a common and valid technique for equation solving, although it takes a little thought to interpret correctly.) May 15, 2015 at 8:29
• @CharlesBailey The rules of equation manipulation are the same, though. Most common algebraic operations applied to equations (addition and subtraction, multiplication and division by nonzeroes) are bidirectional, so the truth of the result is equivalent to the truth of the original statement. Thus if you can use these methods to derive a tautology from the claimed equation, you have indeed made a valid proof. (Usually, one would follow up by reversing it and/or turning it into an equality chain, so you are more likely to see this sort of thing in a student's solution to an algebra problem.) May 15, 2015 at 8:48
• I agree with @Mario Carneiro. You can see the question Why is it that when proving trig identities, one must work both sides independently? and my answer describing "the analytical method for proving an identity. It consists of starting with the identity you want to prove and establish a sequence of identities so that each one is a consequence of the next one." May 15, 2015 at 19:04

$$\cos ^{2}x=1-\sin ^{2}x=(1-\sin x)(1+\sin x)\Longrightarrow \frac{\cos x}{% 1-\sin x}=\frac{1+\sin x}{\cos x}=\sec x+\tan x.$$

# General case

You can employ the tangent half-angle substitution, writing $t:=\tan\frac x2$.

\begin{align*} \frac{\cos(x)}{1-\sin(x)}-\tan(x)&=\sec(x) \\ \frac{\frac{1-t^2}{1+t^2}}{1-\frac{2t}{1+t^2}} - \frac{2t}{1-t^2} &= \frac{1+t^2}{1-t^2} \\ \frac{1-t^2}{1-2t+t^2} - \frac{2t}{1-t^2} &= \frac{1+t^2}{1-t^2} \\ \frac{(1+t)(1-t)}{(1-t)^2} - \frac{2t}{1-t^2} &= \frac{1+t^2}{1-t^2} \\ \frac{(1+t)^2}{(1+t)(1-t)} - \frac{2t}{1-t^2} &= \frac{1+t^2}{1-t^2} \\ (1+t)^2-2t&=1+t^2 \\ 1+2t+t^2-2t&=1+t^2 \end{align*}

# Special cases

If you want to, you can also consider possible special cases. The tangent half-angle substitution I did in my first step doesn't directly capture $x=\pm\pi$, but you get that as $\lim_{t\to\infty}$. Verifying that case explicitely you find

$$\frac{\cos\pi}{1-\sin\pi}-\tan\pi=\frac{1}{1-0}-0=1=\sec\pi$$

In the second step I canceled $1+t^2$ which will be non-zero for real $t$. In the fourth step, I canceled $1-t$ which corresponds to $x=\frac\pi2$. In that case you indeed have a singularity, where pretty much all your terms become undefined. One step later I multiply everything with $1-t^2=(1+t)(1-t)$ which would be illegal for $x=\pm\frac\pi2$. But for $x=-\frac\pi2$ again $\tan(x)$ and $\sec(x)$ become undefined. So the equation holds whenever all the terms it contains are defined, and when all the terms of one side are defined then so are those on the other.

Note that the above does not hold if you are operating in the real projective line $\mathbb R\cup\{\infty\}$. There you have for $x=\frac\pi2$:

$$\frac{\cos\tfrac\pi2}{1-\sin\tfrac\pi2}-\tan\tfrac\pi2=\frac{0}{1-1}-\infty=\frac00-\infty\overset?=\infty=\sec\tfrac\pi2$$

but $\frac00$ is still undefined. So you have a removable singularity in this case.

Split LHS, in terms of $\sin$ and $\cos$

$\frac{\cos(x)}{(1-\sin(x))}-\tan(x)$

$\implies \frac{\cos x}{1-\sin x}-\frac {\sin x}{\cos x}$

$\implies \large\frac{\cos^2 x-\sin x+ sin^2x}{\cos x-\cos x\sin x}$

$\implies \large\frac{1-\sin x}{\cos x(1-\sin x)}$

$\implies \large\frac{1}{\cos x}=\sec x$ =R.H.S

From the definition of sec(x) we have: $$\frac{\cos(x)}{(1-\sin(x))}-\tan(x)=\frac{1}{\cos(x)}$$

I don't like fractions, so lets multiply overall by the denominators: $$(1-\sin(x))\cos(x)$$ and we get $$\cos ^{2}(x) -\tan(x)\cos(x) + \tan(x)\cos(x)\sin(x) = 1-\sin(x)$$ This looks horrible until we realise that tan(x) cos(x) = sin(x), (from the identity tan(x) = sin(x) / cos(x) ). So:

$$\cos ^{2}(x) -\sin(x) + \sin ^{2}(x) = 1-\sin(x)$$ By Pythogoras $$\cos ^{2}(x) + \sin ^{2}(x) = 1$$ and so we have the identity $$1-\sin(x) = 1-\sin(x)$$ and we are done.

Transfer $\tan x$ to right hand side, now it is:

$${\sec x + \tan x } = \dfrac{1+\sin x }{\cos x} =\dfrac{\cos x}{1-\sin x }$$

which is what is left on the left hand side.

Multiply both sides by $(1-\sin(x))\cos(x)$ to get rid of the denominators and verify:

$$\cos^2(x)-(1-\sin(x))\sin(x)=1-\sin(x).$$