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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 `(1/(t_1)+1/(t_2))(1/(t_2)+1/(t_3))=4(1+1/(t2 2))dot`

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Let the points P and Q divide 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 = 2x cot (alpha + beta) - x cot gamma`(i)
From `Delta APC`, we get ltbgt `(x + x) cot theta = x cot beta - x cot gamma`
Dividing (i) by (ii), we get
`(3)/(2) = (2 cot (alpha + beta) - cot gamma)/(cot beta - cot gamma)`
or `3 cot beta - cot gamma = (4 (cot alpha. cot beta -1))/(cot beta + cot alpha)`
or `3 cot^(2) beta - cot beta cot gamma + 3 cot alpha. cot beta - cot alpha. cot gamma = 4 cot alpha. cot beta - 4`
or `4 + 4 cot^(2) beta = cot^(2) beta + cot alpha. cot beta + cot beta. cot gamma + cot gamma. cot alpha`
or `4(1 + cot^(2) beta) = (cot beta + cot alpha) (cot beta + cot gamma)`
or `4(1+(1)/(t_(2)^(2))) = ((1)/(t_(1)) + (1)/(t_(2))) ((1)/(t_(2)) + (1)/(t_(3)))`
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