Sine rule, cosine rule, but tangent rule?

Geometry Level 4

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \text{ } \\ \text{and} \\ \text{ } \\ a ^2 = b^2 + c^2 - 2bc\cos A.$

I think most of us here know the sine rule and cosine rule, which, in case you don't know, are as shown above.

But, did you know there exists a tangent rule as well? Here it is!

Express $\dfrac{\tan \frac{A+B}{2}}{\tan \frac{A-B}{2}}$ in terms of $a$ and $b$ .

Note : Assume all denominators in the fractions in all the answer choices are non-zero.

\frac{a^2+b^2}{a^2-b^2}

\frac{a^2-b^2}{a^2+b^2}

\frac{b-a}{b+a}

\frac{b+a}{b-a}

\frac{b^2-a^2}{b^2+a^2}

\frac{a+b}{a-b}

\frac{a-b}{a+b}

\frac{b^2+a^2}{b^2-a^2}

3 solutions

A Former Brilliant Member
Feb 20, 2016

Since $A ,B,C$ are angles of a triangle,

$A + B + C = \pi \rightarrow A+B = \pi - C$
$\therefore \dfrac{\tan \left(\dfrac{A+B}{2}\right)}{\tan \left(\dfrac{A-B}{2}\right)} = \dfrac{\tan\left(\dfrac{ \pi - C}{2}\right)}{tan\left(\dfrac{A-B}{2}\right)} = \dfrac{\cot \left(\dfrac{ C}{2}\right)}{tan\left(\dfrac{A-B}{2}\right)}$
Using Napier's analogy,
$\tan\left(\dfrac{A-B}{2}\right) = \dfrac{a-b}{a+b}\cdot\cot\left(\dfrac{C}{2}\right)$
$\therefore \dfrac{\cot \left(\dfrac{ C}{2}\right)}{tan\left(\dfrac{A-B}{2}\right)} = \dfrac{a+b}{a-b}$

Rishabh Jain
Feb 5, 2016

Yes its a famous rule named Napier's tangent rule...

@Aloysius Ng : Under what circumstances is it used?

Shaun Leong - 5 years, 4 months ago

For triangles I guess... i.e when $A+B+C=\pi$ and Wolfram says it's also valid for spherical triangles sometimes also called Euler's triangle..

Rishabh Jain - 5 years, 4 months ago

I mean, in a question on triangles what makes it preferable to use it compared to sine rule or cosine rule? I have been taught about it but its usage was not really explained.

Shaun Leong - 5 years, 4 months ago

Well... This rule requires 4 variables so you'll need 3 to solve for the last 1 (or you could get a ratio for a and b if you knew A and B, but you can do that much easily with the Sine Rule)

If you know A, B and a, you can use the Sine Rule If you know A, a and b, you can also use the Sine Rule.

In other words, the Sine Rule is an easier form of this Tangent Rule. In fact, a certain proof for the Tangent rule uses the Sine Rule. Just Google Search "proof of tangent rule" or work it out yourself.

The only way I can imagine that this is useful is if they only gave you a side and tan(A) and tan(B), not A and B, and had to give the exact length of the other side.

So not that useful.

Aloysius Ng - 5 years, 4 months ago

Yashas Ravi
Jul 10, 2019

We can write $\frac{tan(0.5(A+B))}{tan(0.5(A-B))}$ in terms of sine and cosine: $sin(0.5(A+B))*cos(0.5(A-B))$ in the numerator becomes $sin(A)+sin(B)$ by sum to product identities. Similarly, $sin(0.5(A-B))*cos(0.5(A+B))$ in the denominator becomes $sin(A)-sin(B)$ by sum to product. Then, by the law of sines, $b*sin(A)=a*sin(B)$ . By substitution on $sin(A)$ , the $sin(B)$ cancels out in the numerator and denominator, and multiplying the numerator and denominator by $b$ results in $\frac{(a+b)}{(a-b)}$ which is the final answer.

Sine rule, cosine rule, but tangent rule?

3 solutions

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