# Using geometry to prove that $\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)},\sec(\alpha)=\frac{1}{\cos(\alpha)},...$

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Let $$\alpha$$ be the angle $$XOP$$. We know that:

$$\sin(\alpha)=PA$$, and $$\cos(\alpha)=PB$$

$$\tan(\alpha)=PC$$, and $$\cot(\alpha)=PD$$

$$\sec(\alpha)=OC$$, and $$\csc(\alpha)=OD$$

How can we, using geometry, prove that $$PC=\frac{PA}{PB}$$, prove that $$\frac{1}{PA}=OD$$, and so on.

I hope you do not consider this as a silly question. I just believe it is (false) to prove these by recalling the trigonometric functions mentioned above.

Your help would be appreciated. THANKS!

• Formally, the unit of $PC$ is a length whereas the quotient of two lengths has no dimensions. So something is fishy here, you missed a line of length 1 somewhere. May 18, 2022 at 9:31
• @emacsdrivesmenuts $OP=1$, which is the radius of the circle. May 18, 2022 at 10:19

$$\angle CPO=90$$ (tangent), $$\angle OAP=90$$ (AP is altitude)

$$\triangle OPC \sim\triangle OAP\ \ \because AAA$$

$$\ \ \ \angle CPO=\angle OAP$$,

$$\ \ \ \angle OCP=(90-\angle COP)=\angle OPA$$

$$\ \ \ \angle COP$$ is shared

$$\therefore \frac{PA}{OA}=\frac{PC}{OP}$$

But, $$OP=1$$ and $$OA=PB$$

$$\therefore \ PC =\frac{PA}{PB}$$

• Oops. Thank you both for the correction. May 20, 2022 at 11:44