Question

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 $\frac{(\frac{1}{t_1}+\frac{1}{t_2})(\frac{1}{t_2}+\frac{1}{t_3})}{(1+\frac{1}{{t^2}_2})}=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

Let the points $P$ and $Q$ divided the side $BC$ in three equal parts such that $BP=PQ=QC=x$

Also let $\angle BAP=\alpha ,\angle PAQ=\beta ,\angle QAC=\gamma$ and

$\angle AQC=\theta$

From question,

$tan\alpha =t_1,tan\beta =t_2,tan\gamma =t_3$

Applying, $m:n$ rule in triangle $ABC,$ we get $(2x+x)cot\theta =2xcot(\alpha +\beta )-xcot\gamma ...\left(i\right)$

From $\Delta APC$, we get

$(x+x)cot\theta =xcot\beta -xcot\gamma ...\left(ii\right)$

Dividing (i) by (ii) .we get

$\frac{3}{2}=\frac{2cot(\alpha +\beta )-cot\gamma }{cot\beta -cot\gamma }$

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$⟹3cot\beta -cot\gamma =cot\beta +cot\alpha \frac{4(cot\alpha \cdot cot\beta -1)}{cot\beta +cot\alpha }$

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$⟹4(1+co{t}^2\beta )=(cot\beta +cot\alpha )(cot\beta +cot\gamma )$

$⟹4(1+\frac{1}{t_2^2})=(\frac{1}{t_1}+\frac{1}{t_2})(\frac{1}{t_2}+\frac{1}{t_3})$

$⟹\frac{(\frac{1}{t_1}+\frac{1}{t_2})(\frac{1}{t_2}+\frac{1}{t_3})}{(1+\frac{1}{t_2^2})}=4$

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