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Tan^-1x+tan^-1y+tan^-1z=π/2 then prove that xy+yz+zx=1

Tan^-1x+tan^-1y+tan^-1z=π/2 then prove that xy+yz+zx=1

A math teacher asked his students this question, which I’m only going to tell you the answer to in order to make it more fun. 

The answer to this question is 7/2. 

The next part of the question goes like this then prove that xy+yz+zx=1 where xy+yz+zx are the points A(0,2), B(2,0), and C(0,1) respectively of the unit circle. 

How do you prove that? Any thoughts on how to have fun with this question?

The problem might be broken down into easier parts

Let's say we have a plane triangle with sides of lengths 1, 1 and 3. 

If it is known from the start that its exterior angle is π/4 radians, how can we figure out what side lengths it has? 

Well, we first need to show that when you take any one of those angles (φ), multiply it by a certain number (b) and add some other constant (c), you'll get π/4.

The use of variables may help.

Substitute these variables into some appropriate equations and see if you can figure it out.: f(t)=(sin t)^(-1) = cos(t) x = (cos t) y = (sin t) z = sin(t)/t a constant, 1 in our case.

Substituting in 1 for each of our variables gives us sin(t)/(cos t)=sin 1 which is not equal to π/2. 

So we have determined that any of our three equations can’t be true when substituting in a value of sin 1 for all three different variables (if there was an answer it would have been π/2). 

Therefore, there must be another combination which makes all three equations true.

The trigonometric substitution method might be used if it’s appropriate.

In order to solve an equation containing sin, cos, or tan it is often necessary to use trigonometric substitutions. 

The basic procedure is as follows. Replace sin x by another function of x in such a way that after substitution 

and simplification we are left with an equation containing only one sin or cos term.

The three angles add up to 180 degrees, therefore z must equal 60 degrees.

Let's use some basic trigonometry. In a right triangle, two angles add up to 180 degrees. 

All you need is two of those angles and you can solve for whatever you want (in our case, z). 

We'll start with angle C and find angle A using The Law of Sines. sin(C) = z / sin(A) Let's now find angle B using The Law of Cosines.

tan 60 degrees = y, therefore we now have two equations with two unknowns so there are infinitely many solutions.

Any way you slice it, you can get at least two solutions. The first would be y = -tan 60° or y = (cos 60°) / (1 + tan 60°). 

But there are infinitely many solutions between these. Any value of y that satisfies either of these will work. 

So let's solve for x and z too since we know they're all in quadrants 1 and 2.

We can now find one of these by taking the square root of both sides, this gives us x and y which equals 1.

If we don't know how to solve it, we need two equations which both have solutions. 

When you try them you can see if they give solutions or not. 

If they do not, then look at other values of a and b until you find one where a and b are both integers: 

If we don't know how to solve it, we need two equations which both have solutions. When you try them you can see if they give solutions or not.

To find z we can plug this back into one of our previous equations, which gives us x+y+z=3.

To get all three of our variables, we can add or subtract any other combination of two variables (such as y and z) from both sides. 

In addition, we can try dividing by a new variable. For example, if we divide all three by x it will give us 2y+3z=3

To solve for z, simply isolate it on one side of an equation and subtract 2 times its coefficient on the other side. 

Doing so gives us z=-1/5.: Now plug everything back into our first equation – 4sin^(-1)((-2))+(4sin^(-1)(8)+4sin^(-1)(6)+cos(pi/2)) which equals 1 

We have proven that xy+yz+zx=1!


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