The base of a triangle is divide into three equal parts. If $t_{1}, t_{2}, t_{3}$ be the tangents of the angles subtended by these parts at the opposite vertex, then $(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}})$ is equal to :

4 (1 + \frac{1}{t_{2^{2}}})

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4 (1 - \frac{1}{t_{2^{2}}})

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4 (1 + \frac{1}{t_{1^{2}}})

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4 (1 - \frac{1}{t_{1^{2}}})

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Solution

The correct option is A $4 (1 + \frac{1}{t_{2^{2}}})$

Let point $D$ and $E$ divided the base $B C$ into three equal parts i.e., $B D = D E = D C = d$ and
Let $α, β, γ$ be the angles subtended by $B D, D E$ and $E C$ respectively at their opposite vertex.

$t_{1} = tan α, t_{2} = tan β, t_{3} = tan γ$

Now in $Δ A B C$ ,

$B E : E C = 2 d : d = 2 : 1$

From $m - n$ rule, we get

$(2 + 1) cot θ = 2 cot (α + β) - cot γ$

$3 cot θ = 2 cot (α + β) - cot γ$ ---- (1)

Again in $Δ A D C$ ,

$D E : E C = x : x = 1 : 1$

If we apply $m - n$ rule in $Δ A D C$ , we get

$(1 + 1) cot θ = 1. cot β - 1 cot γ$

$2 cot θ = cot β - cot γ$ ----- (2)

From 1 and 2 we get,

$\frac{3 cot θ}{2 cot θ} = \frac{2 cot (α + β) - cot γ}{cot β - cot γ}$

$3 cot β - cot γ = 4 cot (α + β)$

$3 cot β - cot γ = 4 {\frac{cot α . cot β - 1}{cot β + cot α}}$

$3 {cot}^{2} β + 3 cot α cot β - cot β cot α - cot α cot γ = 4 cot α cot β - 4$

$4 + 3 {cot}^{2} β = cot α cot β + cot β cot γ + cot α cot γ$

$4 + 4 {cot}^{2} β = cot α cot β + cot α cot γ + cot β cot γ + {cot}^{2} β$
$4 (1 + {cot}^{2} β) = (cot α + cot β) (cot β + cot α)$

$4 (1 + \frac{1}{{tan}^{2} β}) = (\frac{1}{tan α} + \frac{1}{tan β}) (\frac{1}{tan β} + \frac{1}{tan γ})$

$4 (1 + \frac{1}{t_{2}^{2}}) = (\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}})$

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The base of a triangle is divide into three equal parts. If t1,t2,t3 be the tangents of the angles subtended by these parts at the opposite vertex, then (1t1+1t2)(1t2+1t3) is equal to :

The base of a triangle is divide into three equal parts. If $t_{1}, t_{2}, t_{3}$ be the tangents of the angles subtended by these parts at the opposite vertex, then $(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}})$ is equal to :