Find a value of the angle from the given equation

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To find the angle α from the equation tgα = ((1 + tg1)(1 + tg2) - 2) / ((1 - tg1)(1 - tg2) - 2), the solution involves using the arctangent function. The equation can be simplified to α = arctan{[(1 + t1)(1 + t2) - 2] / [(1 - t1)(1 - t2) - 2]}, where t1 and t2 represent tg1 and tg2. It's confirmed that tg(x) refers to the tangent function, and the desired value of α is approximately 42 degrees. The discussion also suggests using tangent sum and difference rules to derive the exact value.
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Homework Statement


If we are given an equation that equals tg\alpha, and we need to find out how much is \alpha, how would we do it, having in mind the equation bellow?

Homework Equations


tg\alpha=\frac{(1+tg1)(1+tg2)-2}{(1-tg1)(1-tg2)-2}

The Attempt at a Solution


Since i know the answer (i looked it up, and it's 42 degrees), I'm guessing that the right side, when its solved, we should get tg42, but how to do it?

I tried multiplying, but with no luck. Is there an elegant solution.
 
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Chuckster said:

Homework Statement


If we are given an equation that equals tg\alpha, and we need to find out how much is \alpha, how would we do it, having in mind the equation bellow?


Homework Equations


tg\alpha=\frac{(1+tg1)(1+tg2)-2}{(1-tg1)(1-tg2)-2}


The Attempt at a Solution


Since i know the answer (i looked it up, and it's 42 degrees), I'm guessing that the right side, when its solved, we should get tg42, but how to do it?

I tried multiplying, but with no luck. Is there an elegant solution.
The left side is just tg times \alpha? The just divide both sides by tg:

\alpha= \frac{(1+tg1)(1+tg2)-2}{tg((1-tg1)(1-tg2)-2)}

Now, as to what the right side should be, what are t, g, g1, and g2?
 
By tgw, do you mean tan(w) for an angle w? if so, then say it. What you have *written* is a product: t times g times w.

Anyway, if tgw means tan(w), then your equation has a solution alpha = arctan{[(1 + t1)*(1 + t2)-2]/[1 - t1)*(1 - t2) - 2]}, where t1 and t2 are your tg1 and tg2---whatever they may mean. Of course, the solution is not unique because we may add or subtract an integer multiple of pi to the angle and still satisfy the equation.

RGV
 
Ray Vickson said:
By tgw, do you mean tan(w) for an angle w? if so, then say it. What you have *written* is a product: t times g times w.

Anyway, if tgw means tan(w), then your equation has a solution alpha = arctan{[(1 + t1)*(1 + t2)-2]/[1 - t1)*(1 - t2) - 2]}, where t1 and t2 are your tg1 and tg2---whatever they may mean. Of course, the solution is not unique because we may add or subtract an integer multiple of pi to the angle and still satisfy the equation.

RGV

it is tg(alpha), and alpha is in the first quadrant!
also, i need the exact size, in degrees.
 
Will you please answer Ray Vickson's question: does tgx or tg(x) mean tangent of x?

If, indeed, "tgx" or "tg(x)" means tan(x) then Ray Vickson's response is perfectly good answer. What more do you want?
 
HallsofIvy said:
Will you please answer Ray Vickson's question: does tgx or tg(x) mean tangent of x?

If, indeed, "tgx" or "tg(x)" means tan(x) then Ray Vickson's response is perfectly good answer. What more do you want?

I thought i made it clear - yes, it is tangentg of x.

As i said, i already knew that
alpha=arctg of the right side.

But i need how to exactly GET the value of alpha.
I hope you understand where I'm going at?

How much is alpha - in degrees - rather than it is arctg of the expression on the right.
I hope you understood what i meant.
 
Chuckster said:
I thought i made it clear - yes, it is tangentg of x.

As i said, i already knew that
alpha=arctg of the right side.

But i need how to exactly GET the value of alpha.
I hope you understand where I'm going at?

How much is alpha - in degrees - rather than it is arctg of the expression on the right.
I hope you understood what i meant.

Write tan(42)=tan(45-(1+2)) and apply tangent sum and difference rules to it.
 
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