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enter image description here

https://www.desmos.com/calculator/6sdbz1iahd

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!

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  • $\begingroup$ 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. $\endgroup$
    – emacs drives me nuts
    May 18, 2022 at 9:31
  • $\begingroup$ @emacsdrivesmenuts $OP=1$, which is the radius of the circle. $\endgroup$
    – Hussain-Alqatari
    May 18, 2022 at 10:19

1 Answer 1

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$\angle CPO=90$ (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} $

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  • $\begingroup$ Oops. Thank you both for the correction. $\endgroup$
    – robert timmer-arends
    May 20, 2022 at 11:44

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