The base of a triangle is divided into three equal parts If $t_{1}, t_{2}, t_{3}$ are the tangents of the angles subtended by these parts at the opposite vertex,prove that $(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = 4 (1 + \frac{1}{t_{2}^{2}})$

Open in App

Solution

Let the points $P$ and $Q$ divided the side $B C$ in three equal parts such that $B P = P Q = Q C = x$
Also let $∠ B A P = α, ∠ P A Q = β, ∠ Q A C = γ$ and
$∠ A Q C = θ$
From question, $tan α = t_{1}, tan β = t_{2}, tan γ = t_{3}$
Applying, $m : n$ rule in triangle $A B C,$ we get $(2 x + x) cot θ = 2 x cot (α + β) - x cot γ$ ...(i)
From $Δ A P C$ , we get
$(x + x) cot θ = x cot β - x cot γ$ ...(ii)
Dividing (i) by (ii) .we get
$\frac{3}{2} = \frac{2 cot (α + β) - cot γ}{cot β - cot γ}$ $\Rightarrow 3 cot β - cot γ = \frac{4 (cot α \cdot cot β - 1)}{cot β + cot α}$
$\Rightarrow 3 {cot}^{2} β - cot β cot γ + 3 cot α \cdot cot β - cot α \cdot cot γ = 4 cot α \cdot cot β - 4$
$\Rightarrow 4 + 4 {cot}^{2} β = {cot}^{2} β cot α \cdot cot β + cot β cot γ + cot γ \cdot cot α$
$\Rightarrow 4 (1 + {cot}^{2} β) = (cot β + cot α) (c o t β + cot γ)$
$\Rightarrow 4 (1 + \frac{1}{t_{2}^{2}}) = (\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}})$
Hence the result is wrong.

Suggest Corrections

Similar questions

Q. The base of a triangle is divided into

3

equal part. If

t_{1}, t_{2}, t_{3}

be the tangents of the angles subtended by these parts at opposite vertex then value of

\frac{(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}})}{(1 + \frac{1}{{t^{2}}_{2}})} =

Q. 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

Q. 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

The area of the triangle formed by three points $P {a t_{1} t_{2}, a (t_{1} + t_{2})}, Q {a t_{2} t_{3}, a (t_{2} + t_{3})}, R {a t_{3} t_{1}, a (t_{3} + t_{1})}$ is _____.

Q. Three black bodies are such that the maximum intensity of their wavelength are in the ratio

λ_{m_{1}} : λ_{m_{2}} : λ_{m_{3}} = 1 : (21)^{0.5} : (3)^{0.5}

.
Which of these is true for their temperature?

Join BYJU'S Learning Program

The base of a triangle is divided into three equal parts If t1,t2,t3 are the tangents of the angles subtended by these parts at the opposite vertex,prove that (1t1+1t2)(1t2+1t3)=4(1+1t22)

The base of a triangle is divided into three equal parts If $t_{1}, t_{2}, t_{3}$ are the tangents of the angles subtended by these parts at the opposite vertex,prove that $(\frac{1}{t_{1}} + \frac{1}{t_{2}}) (\frac{1}{t_{2}} + \frac{1}{t_{3}}) = 4 (1 + \frac{1}{t_{2}^{2}})$