Preliminary HSC exam question help (1 Viewer)

wiliaamnguuyen · Aug 22, 2015

1. Express tan(A+B) in terms of tanA and tanB
2. Hence show if A, B and C are the vertex angles of a triangle then tan A + tan B + tan C = tanA x tanB x tanC

rand_althor · Aug 22, 2015

InteGrand · Aug 22, 2015

wiliaamnguuyen said:
1. Express tan(A+B) in terms of tanA and tanB
2. Hence show if A, B and C are the vertex angles of a triangle then tan A + tan B + tan C = tanA x tanB x tanC

Alternatively:
$1. $\tan (A+B)=\frac{\tan A + \tan B}{1-\tan A \tan B}$. (So $\tan A + \tan B = \tan(A+B)\cdot (1-\tan A \tan B)$).$

$2. Since $A,B,C$ are angles of a triangle, they satisfy $A+B+C=\pi$ and $A,B,C >0$.$

$Now,$

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

$$ =\tan(A+B)- \tan (A+B)\tan A \tan B + \tan C $ (expanding brackets)$

$$ =\tan(\pi - C)- \tan (\pi - C)\tan A \tan B + \tan C $$

$$=-\tan C+ \tan C\tan A \tan B + \tan C $ (trig. identity: $\tan (\pi - C)=-\tan C$)$

$$=\tan A \tan B\tan C$. \blacksquare$

wiliaamnguuyen · Aug 23, 2015

Can you take me through your thought process? What made you think to do all of that?

InteGrand · Aug 23, 2015

wiliaamnguuyen said:
Can you take me through your thought process? What made you think to do all of that?

$I was sure it would work out by fiddling around with it. The fact that the angles are angles of a triangle made it obvious that we needed to use the fact that $A+B+C=\pi$, because the identity they asked to prove is not true for general angles. This made it clear that using this would require a trigonometric identity for tan.$

Preliminary HSC exam question help (1 Viewer)

wiliaamnguuyen

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rand_althor

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InteGrand

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wiliaamnguuyen

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InteGrand

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