What trigonometric identities can prove this equation to be true?

Question

How can I prove this equation true with trigonometric identities?: $$\frac{\cos(x)}{1-\sin(x)}-\tan(x) = \sec(x)$$

Answer 1

HINT:

You will have to use a little "trick" to simplify the fraction. Try multiplying the fraction by $$\frac{1+\sin x}{1+\sin x}$$ which is the same as $1$, so it will preserve the value of the expression but alter its appearance just enough for you to figure out what to do afterwards.

Answer 2

The only identity needed is $\sin^2(x)+\cos^2(x) = 1$. The other information is just definitions: $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $\sec(x) = \frac1{\cos(x)}$.

First off, I replace everything with sin and cos, and abbreviate them as "s" and "c".

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

becomes

$\frac{c}{1-s}-\frac{s}{c} = \frac1{c} $.

I then clear fractions, remembering that $c \ne 1$ and $s \ne 0$. This becomes

$c^2-s(1-s)=1-s $

or $c^2+s^2-s = 1-s$

Since $c^2+s^2 = 1$, this is $1-s = 1-s$ which, as desired, is an identity.

If you want to do this as a derivation, just go through this with just the left side as a sequence of transformations.

You can even leave it in fraction form:

$\begin{array}\\ \dfrac{c}{1-s}-\dfrac{s}{c} &=\dfrac{c}{1-s}-\dfrac{s}{c}\\ &=\dfrac{c^2-s(1-s)}{c(1-s)}\\ &=\dfrac{c^2-s+s^2}{c(1-s)}\\ &=\dfrac{1-s}{c(1-s)}\\ &=\dfrac{1}{c}\\ \end{array} $

Answer 3

You can write it down as: $$\frac{\cos(x)}{1-\sin(x)}-\tan(x) = \sec(x)$$ $$\frac{\cos(x)}{1-\sin(x)}-\frac{\sin(x)}{\cos(x)} = \frac{1}{\cos(x)}$$ $$\frac{\cos(x)}{1-\sin(x)}=\frac{\sin(x)}{\cos(x)}+\frac{1}{\cos(x)}$$ $$\frac{\cos(x)}{1-\sin(x)}=\frac{1+\sin(x)}{\cos(x)}$$ Changing its form $$\implies (1+\sin x)(1-\sin x) = \cos x \cdot \cos x$$ $$\implies 1^2 - \sin^2x = \cos^2x$$ $$\sin^2x+\cos^2x = 1$$ Which is correct, because it is the Fundamental Theorem of Trigonometry.

Answer 4

Another hint: It is equivalent to $$\frac{\cos^2x}{1-\sin x}-\sin x=1$$ and it boils down to Pythagoras' identity.