MHB How Do You Calculate $ab+bc+ca$ Using Trigonometric Identities?

[anemone]

Here is this week's POTW:

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Compute $ab+bc+ca$ if $a=\tan 15^{\circ},\,b=\tan 25^{\circ},\,c=\tan 50^{\circ}$.

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[anemone]

Congratulations to the following members for their correct solution:):

1. kaliprasad
2. greg1313
3. lfdahl

Solution from kaliprasad:

Spoiler

We have $\tan (A+B+C) =\dfrac{\tan\, A + \tan\, B +\tan\, C - \tan\, A \tan\, B \tan C }{1-\tan\, A \tan\, B - \tan\, B \tan\, C - \tan\, C \tan A } $

If $(A+B+C) = 90^\circ$ then $\tan (A + B + C) = \infty$

So denominator shall be zero

$1-\tan\, A \tan\, B - \tan\, B \tan\, C - \tan\, C \tan A = 0 $

or $\tan\, A \tan\, B + \tan\, B \tan\, C + \tan\, C \tan A = 1 $

as $15 + 25 + 50 = 90$

Therefore $\tan\, 15^\circ \tan\, 25^\circ + \tan\, 25^\circ \tan\, 50^\circ + \tan\, 50^\circ \tan 15^\circ = 1 $

given $a=\tan\,15^\circ$ , $b=\tan\,25^\circ$, $c=\tan\,50^\circ$

we get

$ab+bc+ca = 1$

Solution from greg1313:

Spoiler

$$q=ab+bc+ac$$

$$2q=\tan15^\circ(\tan25^\circ+\tan50^\circ)+\tan25^\circ(\tan15^\circ+\tan50^\circ)+\tan50^\circ(\tan15^\circ+\tan25^\circ)$$

$$\tan15^\circ(\tan25^\circ+\tan50^\circ)=\tan15^\circ\left(\dfrac{\sin25^\circ\cos50^\circ+\sin50^\circ\cos25^\circ}{\cos25^\circ\cos50^\circ}\right)=\dfrac{\sin15^\circ}{\cos25^\circ\cos50^\circ}=\dfrac{\cos75^\circ}{\cos25^\circ\cos50^\circ}$$

and similarly for the other two terms. Hence

$$2q=\dfrac{\cos75^\circ}{\cos25^\circ\cos50^\circ}+\dfrac{\cos65^\circ}{\cos15^\circ\cos50^\circ}+\dfrac{\cos40^\circ}{\cos15^\circ\cos25^\circ}.$$

Now,

$$\dfrac{\cos75^\circ}{\cos25^\circ\cos50^\circ}=\dfrac{\cos25^\circ\cos50^\circ-\sin25^\circ\sin50^\circ}{\cos25^\circ\cos50^\circ}$$

and similarly for the other two terms. Hence

$$2q=3-q\Rightarrow q= 1$$

so we may state $$ab+bc+ac=1.$$