The question is : $\dfrac{\csc x}{\sec x} = \cot x$
After solving a bit I get
$$\frac{\frac1{\sin x}}{\frac1{\cos x}} = \frac{\cos x}{\sin x}$$ $$\frac{\cos x}{\sin x} = \frac{\cos x}{\sin x}$$
Is this right?
$\endgroup$ 41 Answer
$\begingroup$What you did is algebraically correct, but it's only half of the story. Trigonometry is all about triangles (well, it's really about circles, but I digress) which are geometric objects. Taking a geometric approach as well as an algebraic approach would aid a student in developing a deeper understanding of trigonometry.
Consider the following triangle.
Here we have a right triangle with an angle $x$, opposite side $a$, adjacent side $b$, and hypotenuse $c$. Of course, $\cot x=\dfrac{adjacent}{opposite}=\dfrac{b}{a}$
But by multiplying (or dividing) each side by a scaling factor, we can generate a similar triangle; one with all of the same angles, and therefore, all of the same trigonometric ratios, but different side lengths.
Lets try dividing each side by the hypotenuse, $c$.
By looking at the blue triangle, it should be clear now where the identity$\sin^2 x+\cos^2 x=1$ comes from ... the Pythagorean theorem. Moreover, we see that $$\cot x = \frac{adjacent}{opposite}=\frac{\cos x}{\sin x}$$But dividing by $c$ isn't the only transformation that we could have performed. We could have divided by $b$ or $a$, and I leave it to the reader to investigate these further.
One sneaky transformation that we can perform is to multiply the sides of the original triangle by $\dfrac{c}{ab}$ (see diagram).
It is clear that from this triangle that$$\cot x = \frac{adjacent}{opposite}=\frac{\csc x}{\sec x}$$
Perhaps take a handful of problems, and try to do them geometrically, as well as algebraically. It should improve your understanding of trigonometry, and help you to understand all of those (what may seem like) arbitrary definitions.
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