Let's construct a right triangle in the first quadrant of the rectangular coordinate plane. The positive y-axis is one leg (call it y), the positive x-axis is the other leg (call it x), and the hypotenuse is a negatively-sloped line (call it h). h makes an angle with x. Let's call that angle θ.Now, on to trigonometry. We know sinθ=y/h and cosθ=x/h, by the definitions of sine and cosine. And we know that x²+y²=h² by Pythagorean Theorem. With this, we can prove the Pythagorean Identity, which is a major piece of your problem. Take (sinθ)²+(cosθ)²Substitute sinθ=y/h and cosθ=x/h and we get(y/h)²+(x/h)²y²/h²+x²/h²We can add fractions with common denominators (y²+x²)/h²And substitute in h² for (x²+y²) from Pythagorean Theorem, and we geth²/h²1 So for any angle θ, (sinθ)²+(cosθ)²=1Your problem (1+cosθ)(1-cosθ)=sin²θ can be multiplied out into1-cos²θ=sin²θwhich is a rewrite of the Pythagorean Identity. Knowing that sinθ/cosθ=tanθ and 1/cosθ=secθ, we can take the Pythagorean Identity and divide both sides by (cosθ)² to get(tanθ)²+1=(secθ)²If we take your second problem (secθ+tanθ)(secθ-tanθ)=1and multiply out the left side, we get(secθ)²-(tanθ)²=1which is a rewrite of the altered Pythagorean Identity.
There is a difference between identity verification and authenticating identities. Identity verification typically refers to the regulatory process of confirming, for lack of better words, that your customers are who they say they are. Authenticating identities is the process of ensuring that the same user who signs into your product or service is the same person who has been verified.Let’s start with identity verification…The answer to your question is highly dependent on the country you are looking to verify identities in. Let’s assume you are interested in verifying identities in the US (or other western country). My experience leading Google’s identity verification (IDV) team for the past couple years has taught me to expect having a lot less identity data than you want. Given the fact your example involves payments - I’m assuming you are looking to monetize your product / provide some form of financial service. If this is the case, you should expect to be heavily regulated at the state and federal level.Now that that’s out of the way - let’s breakdown the basic IDV flow…Customer enters your product identity verification flowCustomer is prompted with a request for CIP information (Name, Address, DOB, and last 4 of SSN)You can take that information and call a credit API (e.g., Experian, Equifax, TransUnion, IDology)The credit API will respond with a pass/fail responseIf Pass -> CongratulationsIf Fail -> Customer is prompted to entire complete SSN and step 3 happens againDepending on how much you want to spend on verifying each individual, you can request a document (e.g., driver’s license) to help verify the individual.There are many great solutions out there to basically let your app plug-n-play - Jumio, IdentityMind Global, Experian, etc.If you are looking at “at least some level of assurance”, i.e., you want to authenticate customers vs. verify their identities… you should look at integrating an attestation provider (e.g., Google, Facebook, id.me, GlobaliD) in your flow. Your customers can “sign-in” using their digital identities, giving you some level of assurance.For more information on identity - news, media, research, etc. check out One World Identity - The destination for every aspect of identity
I don't know why you'd want to. Nonetheless, there's a general algorithm to verify such identities. Convert everything to sines and cosines. Simplify as you go. Sometimes you'll need the Pythagorean identity [math]\sin^2\theta+\cos^2\theta=1[/math] to finish things up. Your original equation [math]\frac{\csc^2 t}{\cot t} =\csc t \sec t[/math]Since there's a [math]\csc t[/math] on both sides of the equation, let's start by simplifying. The original equation follows from this equation [math]\frac{\csc t}{\cot t} = \sec t[/math]Let's convert to sines and cosines now. That equation is equivalent to [math]\frac{\sin t}{\sin t\cos t} = \frac{1}{\cos t}[/math]Cancel the sines and you get [math]\frac{1}{\cos t} = \frac{1}{\cos t}[/math]which is a valid identity.Since at each stage we reduced the equation to one that implies it, and the final one was valid, so was the original one.