Properties and Solution of Triangles
4.0 Napier's Analogy-Tangent Rule
4.0 Napier's Analogy-Tangent Rule
In a triangle $ABC$ with sides $a=BC$, $b=CA$ and $c=AB$, then
$$\begin{equation} \begin{aligned} (i)\tan (\frac{{B - C}}{2}) = \frac{{b - c}}{{b + c}}\cot \frac{A}{2} \\ (ii)\tan (\frac{{C - A}}{2}) = \frac{{c - a}}{{c + a}}\cot \frac{B}{2} \\ (iii)\tan (\frac{{A - B}}{2}) = \frac{{a - b}}{{a + b}}\cot \frac{C}{2} \\\end{aligned} \end{equation} $$
Proof:
Method $(i)$
LHS $$ \tan \frac{A}{2}\tan (\frac{{B - C}}{2})\cot \frac{A}{2}$$ by using $(\tan \frac{A}{2}\cot \frac{A}{2} = 1)$
$$\begin{equation} \begin{aligned} \tan (\frac{\pi }{2} - \frac{{(B + C)}}{2})\tan (\frac{{B - C}}{2})\cot \frac{A}{2} \\ \cot (\frac{{(B + C)}}{2})\tan (\frac{{B - C}}{2})\cot \frac{A}{2} \\ \frac{{\cos (\frac{{(B + C)}}{2}}}{{\sin (\frac{{(B + C)}}{2}}}\frac{{\sin (\frac{{B - C}}{2})}}{{\cos (\frac{{B - C}}{2})}}\cot \frac{A}{2} \\ \frac{{2\cos (\frac{{(B + C)}}{2}}}{{2\sin (\frac{{(B + C)}}{2}}}\frac{{\sin (\frac{{B - C}}{2})}}{{\cos (\frac{{B - C}}{2})}}\cot \frac{A}{2} \\ \frac{{\sin B - \sin C}}{{\sin B + \sin C}}\cot \frac{A}{2} \\\end{aligned} \end{equation} $$
By using sine rule $$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = k$$
$$\begin{equation} \begin{aligned} \frac{{\frac{b}{k} - \frac{c}{k}}}{{\frac{b}{k} + \frac{c}{k}}}\cot \frac{A}{2} \\ \frac{{b - c}}{{b + c}}\cot \frac{A}{2} \\\end{aligned} \end{equation} $$ which is equal to RHS
Method $(ii)$
We know that $$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = k$$
Consider $$\begin{equation} \begin{aligned} \frac{{\frac{b}{k} - \frac{c}{k}}}{{\frac{b}{k} + \frac{c}{k}}}\cot \frac{A}{2} \\ \frac{{b - c}}{{b + c}} = \frac{{k\sin B - k\sin C}}{{k\sin B + k\sin C}} \\ \frac{{b - c}}{{b + c}} = \frac{{k(\sin B - \sin C)}}{{k(\sin B + \sin C)}} \\ \frac{{b - c}}{{b + c}} = \frac{{(\sin B - \sin C)}}{{(\sin B + \sin C)}} \\ \frac{{b - c}}{{b + c}} = \frac{{2\cos (\frac{{(B + C)}}{2}}}{{2\sin (\frac{{(B + C)}}{2}}}\frac{{\sin (\frac{{B - C}}{2})}}{{\cos (\frac{{B - C}}{2})}} \\ \frac{{b - c}}{{b + c}} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{A}{2}}}\tan (\frac{{B - C}}{2}) \\ \frac{{b - c}}{{b + c}} = \tan \frac{A}{2}\tan (\frac{{B - C}}{2}) \\ \frac{{b - c}}{{b + c}}\cot \frac{A}{2} = \tan (\frac{{B - C}}{2}) \\\end{aligned} \end{equation} $$
Question 10. Find the unknowns of triangle $ABC$ in which $a = \sqrt 3 + 1,b = \sqrt 3 - 1,\angle C = {60^0}$.
Solution: $$\begin{equation} \begin{aligned} a = \sqrt 3 + 1,b = \sqrt 3 - 1,\angle C = {60^0} \\ A + B + C = {180^0} \\ A + B = {120^0}...(i) \\\end{aligned} \end{equation} $$
From napier's analogy, we have $$\begin{equation} \begin{aligned} \tan (\frac{{A - B}}{2}) = \frac{{a - b}}{{a + b}}\cot \frac{C}{2} \\ \tan (\frac{{A - B}}{2}) = \frac{{(\sqrt 3 + 1) - (\sqrt 3 - 1)}}{{(\sqrt 3 + 1) + (\sqrt 3 - 1)}}\cot {30^0} \\ \tan (\frac{{A - B}}{2}) = \frac{2}{{2\sqrt 3 }}\cot {30^0} \\ \tan (\frac{{A - B}}{2}) = 1 \\ \frac{{A - B}}{2} = \frac{\pi }{4} \\ A - B = {90^0}...(ii) \\\end{aligned} \end{equation} $$
Solving $(i)$ and $(ii)$, we get $$\begin{equation} \begin{aligned} A = {105^0} \\ B = {15^0} \\\end{aligned} \end{equation} $$
Now from the sine rule $$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = k$$
$$c = \frac{{a\sin C}}{{\sin A}} = \frac{{(\sqrt 3 + 1)\sin {{60}^0}}}{{\sin {{105}^0}}}$$
$$\sin {105^0} = \sin ({60^0} + {45^0})$$ Now using $$\begin{equation} \begin{aligned} \sin (A + B) = \sin A\cos B + \cos A\sin B \\ \sin {105^0} = \sin {60^0}\cos {45^0} + \cos {60^0}\sin {45^0} \\ \sin {105^0} = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} \\ \sin {105^0} = \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \\\end{aligned} \end{equation} $$
So putting the value of $\sin {105^0}$ ,we have$$\begin{equation} \begin{aligned} c = \frac{{a\sin C}}{{\sin A}} = \frac{{(\sqrt 3 + 1)\frac{{\sqrt 3 }}{2}}}{{\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}}} \\ c = \sqrt 6 \\ c = \sqrt 6 ,A = {105^0},B = {15^0} \\\end{aligned} \end{equation} $$
Question 11. In a triangle $ABC$, we define $x = \tan (\frac{{B - C}}{2})\tan \frac{A}{2},y = \tan (\frac{{C - A}}{2})\tan \frac{B}{2},z = \tan (\frac{{A - B}}{2})\tan \frac{C}{2}$, then show that $$x+y+z=-xyz$$
Solution: Using napier's analogy $$\begin{equation} \begin{aligned} (i)\tan (\frac{{B - C}}{2}) = \frac{{b - c}}{{b + c}}\cot \frac{A}{2} \\ (ii)\tan (\frac{{C - A}}{2}) = \frac{{c - a}}{{c + a}}\cot \frac{B}{2} \\ (iii)\tan (\frac{{A - B}}{2}) = \frac{{a - b}}{{a + b}}\cot \frac{C}{2} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} x + y + z = (\frac{{b - c}}{{b + c}}\cot \frac{A}{2}\tan \frac{A}{2}) + (\frac{{c - a}}{{c + a}})\cot \frac{B}{2}\tan \frac{B}{2} + \frac{{a - b}}{{a + b}}\cot \frac{C}{2}\tan \frac{C}{2} \\ x + y + z = \frac{{b - c}}{{b + c}} + \frac{{c - a}}{{c + a}} + \frac{{a - b}}{{a + b}} \\ x + y + z = \frac{{(b - c)(c + a)(a + b) + (c - a)(b + c)(a + b) + (a - b)(b + c)(c + a)}}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(a + b)\{ bc + ab - {c^2} - ac + bc + {c^2} - ab - ac\} + (a - b)(b + c)(c + a)}}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(a + b)(2bc - 2ac) + (a - b)(b + c)(c + a)}}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(a + b)2c(b - a) + (a - b)(b + c)(c + a)}}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(b - a)\{ 2c(a + b) - (b + c)(c + a)\} }}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(b - a)\{ 2ca + 2cb - bc - ba - {c^2} - ca\} }}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(b - a)\{ ca + cb - ba - {c^2}\} }}{{(b + c)(c + a)(a + b)}}, \\ x + y + z = \frac{{(b - a)\{ c(a - c) - b(a - c)\} }}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{(b - a)(c - b)(a - c)}}{{(b + c)(c + a)(a + b)}} \\ x + y + z = \frac{{ - (a - b)(b - c)(c - a)}}{{(b + c)(c + a)(a + b)}} \\\end{aligned} \end{equation} $$
Now considering RHS, we get $$\begin{equation} \begin{aligned} - xyz = - (\frac{{b - c}}{{b + c}}\cot \frac{A}{2}\tan \frac{A}{2}) \times (\frac{{c - a}}{{c + a}})\cot \frac{B}{2}\tan \frac{B}{2} \times \frac{{a - b}}{{a + b}}\cot \frac{C}{2}\tan \frac{C}{2} \\ - xyz = \frac{{b - c}}{{b + c}} \times \frac{{c - a}}{{c + a}} \times \frac{{a - b}}{{a + b}} \\ - xyz = \frac{{ - (a - b)(b - c)(c - a)}}{{(b + c)(c + a)(a + b)}} \\\end{aligned} \end{equation} $$
