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Just looking for help on this particular question:

Given that $\tan (x - y) = k$ and $\tan(x) = 1$, express $\tan(y)$ in terms of $k$**

So far I've used the trig identity: $\tan(A±B) = (\tan(A)±\tan(B)) / (1 ∓ \tan(A)\tan(B))$

This got me to $(1-\tan(y))/(1+\tan(y))=k$, but I'm unsure where to go from here and struggle with solving these types of equations. From here, I tried to multiply the fraction by $(1-\tan(y))/(1-\tan(y))$, but this didn't help.

Thanks! Toby.

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    $\begingroup$ Multiply by $1+\tan y$ on both sides, and solve for $\tan y$, it should be ok $\endgroup$ – H. Potter Jun 2 '16 at 12:55
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Hint: Assuming your equation $$\frac{1-\tan y}{1 + \tan y} = k$$ is correct, note that you can write $$\frac{1-\tan y}{1 + \tan y} = \frac{2}{1 + \tan y}-1$$ Replace the LHS of your equation with this. You should see how to isolate $\tan y$ now.

Addendum: Completing since another answer has been accepted. $$\frac{1-\tan y}{1 + \tan y} = k$$ $$\frac{2}{1 + \tan y}-1 = k\tag{rewrite LHS}$$ This is a very nice form because now "$\tan y$" appears in only one place. We can immediately solve for "$\tan y$": $$\frac{2}{1 + \tan y}= 1+k\tag{add $1$}$$ $$\frac{1}{1 + \tan y} = \frac{1+k}{2}\tag{div. by $2$}$$ $$1 + \tan y = \frac{2}{1+k}\tag{invert}$$ $$\tan y = \frac{2}{1+k}-1\tag{subtr. $1$}$$ It may be a little nicer to combine these terms: $$\tan y = \frac{1-k}{1+k}\tag{rewrite RHS}$$ To sum up, our first step put the equation in a nice form. It's nice because you can immediately see that the LHS was created from "$\tan y$" by successively performing the steps "add $1$, invert, multiply by $2$, subtract $1$". So we just need to undo these steps in reverse order on both sides to obtain "$\tan y$" on the LHS, and the RHS becomes whatever it becomes since it's along for the ride.

Addendum 2: The "trick" to nicing up the original form is to note that you can do this sort of thing: $$\frac{a+b\;\boxed{X}}{c+d\;\boxed{X}}=\frac{a+\frac{b}{d}(dX)}{c+dX}=\frac{a -\frac{b}{d}(c)+\frac{b}{d}(c+dX)}{c+dX}$$ $$=\frac{\left(a -\frac{bc}{d}\right)}{c+d\;\boxed{X}} +\left(\frac{b}{d}\right)$$ All the stuff in parentheses is just some number, and now the "$X$" is by itself in one spot.

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  • $\begingroup$ In the second step, how did you get (1 − tan(y)) / (1 + tan(y)) = 2 / (1 + tan(y)) −1 ? $\endgroup$ – Toby Mellor Jun 2 '16 at 20:38
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    $\begingroup$ That's the "trick" I allude to above. Let's use "$t$" as shorthand for "$\tan y$". The denominator is $1+t$, so write the numerator as $$1-t=1-\overbrace{(1+t -1)}^{t} = 1-(1+t)+1=2-(1+t)$$ Then the fraction is $$\frac{2-(1+t)}{1+t}=\frac{2}{1+t} - \frac{1+t}{1+t}=\frac{2}{1+t}-1$$ The point is to end up with a constant multiple of denominator in the numerator, so it cancels leaving only a number. The rest of the numerator is also just a number. All of the "$t$" was written as a number plus some multiple of $1+t$ (the multiplier of $(1+t)$ is the original multiplier of $t$ in the numerator). $\endgroup$ – MPW Jun 2 '16 at 21:09
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    $\begingroup$ (continued) This is really just doing polynomial division $(1-t)\div (1+t)$, which you would write as $(-t + 1)\div (t+1)$. This gives a quotient of $-1$ with a remainder of $2$, that is, $-1 + \frac{2}{t+1}$. Maybe that's the easiest way to think of it. $\endgroup$ – MPW Jun 2 '16 at 21:19
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Hint:

Use the identity:

$$\tan(x-y) =\frac{\tan x - \tan y}{1 + \tan x\tan y} = k$$

Now factorise $\tan y$ out

$$k + k \tan y = 1 - \tan y $$

$$ \tan y (k+1) = 1 - k$$

You can finish it off from here

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