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Given in a triangle ABC ,if $\angle A =45^\circ $, find all possible values of $\tan(B)\tan(C)$

Now $B+C= \frac{3\pi}{4}$

Taking tan on both sides and cross multiplying we get

$-1+\tan(B)\tan(C)=\tan(B)+\tan(C)$

Let $\tan(B)\tan(C)=x$. Now basically what i intended to do here is that i want to find out values of $\tan(A),\tan(B)$ in terms of $x$. However i am not sure how?

Thanks

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    I'm not sure where you are getting $-1+\tan(B)\tan(C) = \tan(A)+\tan(B)$ taking $\tan$ of both sides and multiplying what? Also you have $B+C\ge 3\pi/4$ not equal (and remember you also have $B+C < \pi$) – 2017-02-07
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    @spaceisdarkgreen: OP is referring to the tangent of sums expression: $\tan(B+C) = \frac{\tan B + \tan C}{1+\tan B \tan C}$. Since $B+C = \frac{3\pi}{4}$, both sides of that expression are equal to $-1$. – 2017-02-07
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    @BrianTung Ahh I see. I also must have flubbed the angle sign for a "Less than". That makes it an easier problem. – 2017-02-07

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You can use the identiy$$ \tan(B+C) = \frac{\tan(B) +\tan(C)}{1+\tan(B)\tan(C)}$$ so that $$\tan(B)\tan(C) = \frac{\tan(B)+\tan(C)}{\tan(B+C)} -1$$ As you said, $B+C=3\pi/4$ so $$ \tan(B)\tan(C) = -(\tan(B)+\tan(3\pi/4-B)+1).$$

Plotting this from $B=0$ to $B=3\pi/4$ shows that $\tan(B)\tan(C)$ starts from zero, increases to infinity as $B\to \pi/4^-,$ so it can take any non-negative value.

For negative values, observe that the when $\pi/4

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    Are you sure about $\tan(BC)$? It should be $\tan B\tan C$.. – 2017-02-07
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    @MattG88 Yuck. Having all sorts of problems with this one. – 2017-02-07
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    @spaceisdarkgreen is not there a simple solution to this problem. i know no differentiation or calculus yet – 2017-02-07
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    @SophieClad I know of no way other than by analyzing some functions graphically (See the approach from the other answerer). I could easily be missing something though. Note that you don't actually need any calculus to do this. If you plot the function, you'll see that there's a peak in the middle that is the largest negative value the function gets to. Don't need calculus to know where the middle is. – 2017-02-07