Chapter 8.2, Problem 61AYU

To determine

To drive the law of tangent using Mollweide’s Formula.

Answer to Problem 61AYU

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

Explanation of Solution

Given information:

To drive the law of tangent using Mollweide’s Formula,

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

Calculation:

Let us consider the $\frac{a-b}{a+b}$ ,divide numerator and denominator by $c$ we get,

  $\frac{a-b}{a+b}=\frac{\frac{a-b}{c}}{\frac{a+b}{c}}$

Use Mollweide’s Formula on,

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

  $\begin{array}{l}\frac{\frac{a-b}{c}=\frac{\sin \left[\frac{1}{2}\left(A-B\right)\right]}{\cos \frac{C}{2}}}{\frac{a+b}{c}=\frac{\cos \left[\frac{1}{2}\left(A-B\right)\right]}{\sin \frac{C}{2}}}\\ \frac{\frac{\sin \left[\frac{1}{2}\left(A-B\right)\right]}{\cos \frac{C}{2}}}{\frac{\cos \left[\frac{1}{2}\left(A-B\right)\right]}{\sin \frac{C}{2}}}=\frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\cot \frac{C}{2}}\\ \frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\cot \frac{C}{2}}=\frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)}\left[\therefore \cot \frac{C}{2}=\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)\right]\end{array}$

If $A,B,C$ are angle of triangle than $A+B+C=\pi ,A+B=\pi -C$

  $\frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\tan \left[\frac{1}{2}\left(\frac{\pi }{2}-\frac{C}{2}\right)\right]}=\frac{\tan \left[\frac{1}{2}\left(A-B\right)\right]}{\tan \left[\frac{1}{2}\left(A+B\right)\right]}$

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