Let `O` be a point inside a triangle `A B C` such that `/_O A B=/_O B C=/_O C A=omega` , then show that: `cotomega=cotA+cotB+cot C` `Cos e c^2omega=cos e c^2A+cos e c^2B+cos e c^2C`

Options

A. `tan^(2) theta`
B. `cot^(2) theta`
C. `tan theta`
D. `cot theta`

Answer

D

Explanation

Applying sine rule in `DeltaAOB`, we have
`(OA)/(sin angleABO) = (AB)/(sin angle AOB)`
or `OA = (c sin angleABO)/(sin angleAOB) = (c sin (B - theta))/(sin B)`...(i)
`[ :' angle ABO = B - theta, angle AOB = 180^(@) - theta - angleABO = 180^(@) -B]`
Again in `DeltaAOC`, we have
`(OA)/(sin angleACO) = (AC)/(sin angleAOC)`
`rArr OA = (b sin angleACO)/(sin angleAOC) = (b sin theta)/(sin A)`
`[ :' angleOAC = A - theta, angleAOC = 180^(@) - theta - angleOAC = 180^(@)]`
From Eqs. (i) and (ii), we have
`(c sin (B - theta))/(sin B) = (b sin theta)/(sin A)`
or `c sin A (B - theta) = b sin theta sin B`
`= b sin theta sin (A +C)`
or `2R sin C sin A (sin B cos theta - cos B sin theta)`
`= 2R sin B sin theta (sin A cos C + cos A sin C)`
Dividing both sides by `2R sin theta sin A sin B sin C`, we get
`cot theta - cot B = cot C + cot A`
or `cot theta = cot A + cot B + cot C`
Squaring both sides, we have
`cot^(2) theta = cot^(2) A + cot^(2) B + cot^(2)C + 2(cotA cot B + cot B cot C + cot C cot A)`
or `cosec^(2) theta - 1 = (cosec^(2) A -1) + (cosec^(2) B -1) + (cosec^(2) C -1) + 2`
[since in `DeltaABC, cot A cot B + cot B cot C + cot C cot A = 1`]
or `cosec^(2) theta = cosec^(2) A + cosec^(2) B + cosec^(2)C`
Area of triangle ABC,
`Delta = Delta_(1) + Delta_(2) + Delta_(3)`
`=(1)/(2) [a OB + b OC + c OA] sin theta`
`=(1)/(4) tan theta [2 a OB cos theta + 2b OC cos theta+ 2c OA cos theta]`
`=(1)/(4) tan theta [(a^(2) + x^(2) -y^(2)) + (b^(2) + y^(2) - z^(2)) + (c^(2) + z^(2) - x^(2)]`
`= (1)/(4) tan theta [a^(2) + b^(2) + c^(2)]`