Question

Law of Tangents For any triangle, derive the Law of Tangents.

$\frac{a-b}{a+b}=\frac{\tan \left[{\displaystyle \frac{1}{2}}\left(A-B\right)\right]}{\tan \left[\frac{1}{2}\left(A+B\right)\right]}$

Short answer

It is proved that$\frac{a-b}{a+b}=\frac{\tan \left[{\displaystyle \frac{1}{2}\left(A-B\right)}\right]}{\tan \left[\frac{1}{2}\left(A+B\right)\right]}$

Step-by-step solution

Step 1. Given Information

Mollweide's formula $$\begin{gathered} \frac{a+b}{c}=\frac{\cos \left[{\displaystyle \frac{1}{2}}\left(A-B\right)\right]}{\sin \left({\displaystyle \frac{1}{2}}C\right)}-----\left(1\right) \\ \frac{a-b}{c}=\frac{\sin \left[{\displaystyle \frac{1}{2}\left(A-B\right)}\right]}{\cos \left({\displaystyle \frac{1}{2}C}\right)}------\left(2\right) \end{gathered}$$

Sum of angles in a triangle$A+B+c={180}^{\circ }$

Step 2. Multiply equation (2) with the reciprocal of equation (1)

$$\begin{gathered} \frac{a-b}{c}\times \frac{c}{a+b}=\frac{\sin \left[{\displaystyle \frac{1}{2}\left(A-B\right)}\right]}{\cos \left({\displaystyle \frac{1}{2}C}\right)}\times \frac{\sin \left(\frac{1}{2}C\right)}{\cos \left[\frac{1}{2}\left(A-B\right)\right]} \\ \frac{a-b}{a+b}=\frac{\sin \left[\frac{1}{2}\left(A-B\right)\right]}{\cos \left[\frac{1}{2}\left(A-B\right)\right]}\times \frac{\sin \left(\frac{1}{2}C\right)}{\cos \left(\frac{1}{2}C\right)} \end{gathered}$$

Step 3. Finding C2

$$\begin{gathered} A+B+C=180 \\ C=180-A-B \\ \frac{C}{2}=90-\left(\frac{A+B}{2}\right) \end{gathered}$$

Step 4. Substitute C2=90∘-(A+B)2 into sinC2

$\sin \frac{C}{2}=\sin \left(90-\frac{A+B}{2}\right)=\cos \left(\frac{A+B}{2}\right)$

Step 5. Substitute C2=90∘-(A+B)2into cosC2.

Step 6. Plug sinA+B2 & cosA+B2into a-ba+b

$$\begin{gathered} \frac{a-b}{a+b}=\frac{\sin {\displaystyle \frac{A-B}{2}}}{\cos \frac{A-B}{2}}\times \frac{\cos \frac{A+B}{2}}{\sin \frac{A+B}{2}} \\ \frac{a-b}{a+b}=\tan \left(\frac{A-B}{2}\right)\times \frac{1}{\tan \left({\displaystyle \frac{A+B}{2}}\right)} \\ \frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)} \end{gathered}$$