In `Delta ABC, angle B=(pi)/(4),angle C=(pi)/(6)`. D is a point on BC which divides it in the ratio `1 : 3, angle DAB=beta`, then

Options

A. `(sec.(pi)/(6))AB+cot beta=cot((pi)/(6))AC+(sqrt(3)-5)`
B. `(sec.(pi)/(4)AB cot beta=cot((pi)/(4))AC(4sqrt(3)-5)`
C. `(sec.(pi)/(4))AB+cot beta=cot((pi)/(6))AC+(4sqrt(3)+5)`
D. `(sec.(pi)/(6))AB cot beta=cot((pi)/(4))AC(sqrt(3)+5)`

Answer

B

Explanation

In `Delta ABD`,
`(sin beta)/((1)/(4)BC)=(sin. (pi)/(4))/(AD)` and `(sin((7pi)/(12)-beta))/((3)/(4)BC)=(sin.(pi)/(6))/(AD)`
Dividing, we get
`(sin((7pi)/(12)-beta))/(sin beta)=(3sqrt(2))/(2)`
`rArr (sin((5pi)/(12)+beta))/(sin beta)=(3sqrt(2))/(2)`
`rArr cot beta = 4 sqrt(3)-5`
In `Delta ABC, (AB)/(sin.(pi)/(6))=(AC)/(sin.(pi)/(4))`
`rArr sqrt(2)AB=AC`
`rArr (sec.(pi)/(4))AB=(cot.(pi)/(4))AC`
Multiplying equations (1) and (2), we get
`(sec.(pi)/(4))AC cot beta = cot.((pi)/(4))AC(4-sqrt(3)-5)`