The base of triangle is divided into three equal parts. If $t_{1}, t_{2}, t_{3}$ be the tangents of the angle subtended by these parts at the opposite vertices, then $(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = k (1 + \frac{1}{t_{2}^{2}})$ . Find the value of $k$ .

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Solution

To find: Value of K
In $△ A B E, (1 + 1) cot α = cot θ_{1} - cot θ_{2} . . . (1)$
In $△ A B C,$ consider $A D$ as the median and apply m-n theorem
$(1 + 2) cot α = 1 cot θ_{1} - 2 cot {(θ}_{2} + θ_{3}) . . . (2)$
Dividing the above equations
$\frac{2}{3} = \frac{cot θ_{1} - cot θ_{2}}{cot θ_{1} - 2 cot {(θ}_{2} + θ_{3})}$
$2 cot θ_{1} - 4 cot {(θ}_{2} + θ_{3}) = 3 cot θ_{1} - 3 cot θ_{2}$
$3 cot θ_{2} = cot θ_{1} + 4 cot {(θ}_{2} + θ_{3})$
$\frac{3}{t_{2}} = \frac{1}{t_{1}} + 4 \times \frac{\frac{1}{t_{2} t_{3}} - 1}{\frac{1}{t_{2}} + \frac{1}{t_{3}}}$
${t_{2}}^{2} + t_{2} t_{3} + t_{1} t_{3} + t_{1} t_{2} = 4 t_{1} t_{3} + 4 {t_{1} t}_{2}^{2} t_{3}$
$t_{2} {(t_{2} + t}_{3}) + t_{1} {(t_{2} + t}_{3} {) = 4 t}_{1} t_{3} (1 + {t_{2}}^{2})$
$\frac{{(t_{1} + t}_{2})}{{t_{1} t}_{2}} \times \frac{{(t_{2} + t}_{3})}{{t_{2} t}_{3}} = 4 \times \frac{1 + {t_{2}}^{2}}{{t_{2}}^{2}}$
$(\frac{1}{t_{1}} + \frac{1}{t_{2}}) \times (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = 4 \times (1 + \frac{1}{{t_{2}}^{2}})$
$∴ k = 4$

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