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. The relationship between $t_{1}, t_{2}, t_{3}$ is given by the following equation
$(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = k (1 + \frac{1}{t_{2}^{2}})$ . Then the value of $k$ is

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Solution

Considering $A D$ as the median and applying $m - n$ theorem
in $△ A B E,$ we have: $(1 + 1) cot α = cot θ_{1} - cot θ_{2} \dots (i)$
$(1 + 2) cot α = 1 cot θ_{1} - 2 cot (θ_{2} + θ_{3}) \cdot (i i)) Diving the above equations \frac{2}{3} = \frac{cot θ_{1} - cot θ_{2}}{cot θ_{1} - 2 cot (θ_{2} + θ_{3})} 2 cot θ_{1} - 4 cot (θ_{2} + θ_{3}) = 3 cot θ_{1} - 3 cot θ_{2} 3 cot θ_{2} = cot θ_{1} + 4 cot (θ_{2} + θ_{3}) \frac{3}{t_{2}} = \frac{1}{t_{1}} + 4 \times \frac{\frac{1}{t_{2} t_{3}} - 1}{\frac{1}{t_{2}} + \frac{1}{t_{3}}}$
On simplifying the above expression, we have:
$t_{2} (t_{2} + t_{3}) + t_{1} (t_{2} + t_{3}) = 4 t_{1} t_{3} (1 + t_{2}^{2}) \frac{(t_{1} + t_{2})}{t_{1} t_{2}} \times \frac{(t_{2} + t_{3})}{t_{2} t_{3}} = 4 \times \frac{1 + t_{2}^{2}}{t_{2}^{2}} (\frac{1}{t_{1}} + \frac{1}{t_{2}}) \times (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = 4 \times (1 + \frac{1}{t_{2}^{2}})$
$∴ k = 4$

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Pair of Tangents from an External Point

Standard XII Mathematics

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The base of triangle is divided into three equal parts. If t1,t2,t3 be the tangents of the angle subtended by these parts at the opposite vertices. The relationship between t1,t2,t3 is given by the following equation (1t1+1t2)(1t2+1t3)=k(1+1t22). Then the value of k is