Question is: verify the identity:
$$
\frac{\cos x}{1 – \sin x} – \tan x = \sec x.
$$
How do I show that the left side equals the right?
I changed $\tan x$ into $\sin x/\cos x$ but didn't get anywhere.
Please help.
[Math] How to show that $ \frac{\cos x}{1 – \sin x} – \tan x = \sec x$
trigonometry
Related Solutions
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, $$\triangle SPR \sim \triangle OQR \implies \frac{|\overline{SP}|}{|\overline{SR}|} = \frac{|\overline{OQ}|}{|\overline{OR}|} \implies \frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta+\tan\theta}{1}$$
One way we can prove the identity false is as follows:
$$\begin {align} \dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\ 2 = \dfrac {1}{1-2\cos^2 2A} \\ 2 (1-2\cos^2 2A) = 1 \\ 2 - 4\cos^2 2A = 1 \\ - 4\cos^2 2A = \dfrac {1}{2} \\ \cos^2 2A = -\dfrac {1}{8} \end {align}$$
Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.
Best Answer
$$\frac{\cos x}{1-\sin x}-\tan x=\frac{\cos x}{1-\sin x}-\frac{\sin x}{\cos x}=$$
$$=\frac{\cos^2x-\sin x+\sin^2x}{\cos x(1-\sin x)}=\frac{1-\sin x}{\cos x (1-\sin x)}=\ldots$$