In a `DeltaABC,(a+c)/(a-c)tanB/2` is equal to

Options

A. `tan(B/2+C)`
B. `tan(B+C/2)`
C. `cot(B/2+C)`
D. none of these

Explanation

Using sine rule, we have
`(a+c)/(a-c)tanB/2=(sinA+sinC)/(sinA-sinC)tanB/2`
`rArr(a+c)/(a-c)tanB/2=(2sin(A+C)/2cos(A-C)/2)/(2sin(A-C)/2cos(A+C)/2)tanB/2`
`rArr(a+c)/(a-c)tanB/2=(cosB/2cos(A-C)/2)/(sin(A-C)/2sinB/2)tanB/2`
`rArr(a+c)/(a-c)tanB/2=(cosB/2cos(A-C)/2)/(sin(A-C)/2sinB/2).tanB/2`
`rArr(a+c)/(a-c)tanB/2=cot((A-C)/2)`
`rArr(a+c)/(a-c)tanB/2=cot((A-C)/2)`
`rArr(a+c)/(a-c)tanB/2=cot((pi-B-C-C)/2)[becauseA+B+C=pi]`
`rArr(a+c)/(a-c)tanB/2=tan(B/2+C)`