Question

state and prove
napier formula

Answer 1

Answer: Napier formula also known as Laws of tangent is given by 3 equations. These equation are as follows:-

1. $cot(\frac{A}{2})\frac{b-c}{b+c} = tan\frac{B-C}{2}$
2. $cot(\frac{B}{2})\frac{c-a}{c+a} = tan\frac{C-A}{2}$
3. $cot(\frac{C}{2})\frac{a-b}{a+b} = tan\frac{A-B}{2}$

Step-by-step explanation:

Step 1: Prerequisite

Consider a triangle ABC with angles A+B+C = π

$\frac{B+C}{2} = \frac{\pi}{2}-\frac{A}{2}$

a, b, and c be respective lengths of sides AB, BC, AC

Step 2: Proof of 1st equation

In any triangle ABC

$\frac{b}{sin B} = \frac{c}{sin C}\\ \frac{b}{c} =\frac{sin B}{sin C} \\\frac{b-c}{b+c} = \frac{sinB - sin C}{sin B + sin C} [Applying Dividendo and Componendo]\\\frac{b-c}{b+c} = \frac{2cos\frac{B+C}{2}sin\frac{B-C}{2} }{2sin\frac{B+C}{2}cos\frac{B-C}{2}}\\\frac{b-c}{b+c} = cot\frac{B+C}{2}tan\frac{B-C}{2}\\\frac{b-c}{b+c}= cot({\frac{\pi }{2}-\frac{A}{2})tan\frac{B-C}{2}\\ \frac{b-c}{b+c} = \frac{tan\frac{B-C}{2}}{cot\frac{A}{2}}$

$cot(\frac{A}{2})\frac{b-c}{b+c} = tan\frac{B-C}{2}$

Similarly using the above technique other 2 equations can also be proven.