If `(a-b)sin (theta+phi)=(a+b)sin(theta-phi)` and `a tan. (theta)/(2)-b tan.(phi)/(2)=c`, then

Options

A. `b tan phi = a tan theta`
B. `a tan phi = b tan theta`
C. `sin phi=(2bc)/(a^(2)-b^(2)-c^(2))`
D. `sin theta=(2ac)/(a^(2)-b^(2)+c^(2))`

Answer

B, C, D

Explanation

`(a-b)sin(theta+phi)=(a+b)sin(theta-phi)`
`rArr (sin(theta+phi))/(sin(theta-phi))=(a+b)/(a-b)`
`rArr (sin(theta + phi)+sin(theta-phi))/(sin(theta+phi)-sin(theta-phi))=(2a)/(2b)`
`rArr (2sin theta cos phi)/(2 cos theta sin phi)=(a)/(b)=(tan theta)/(tan phi)=(a)/(b)`
`rArr b tan theta = a tan phi`
`therefore` (b) is true
`rArr (2a tan.(phi)/(2))/(1-"tan"^(2)(phi)/(2))=(2b tan.(theta)/(2))/(1-"tan"^(2)(theta)/(2))`
`rArr (a((a tan.(theta)/(2)-c)/(b)))/(1-((a tan.(theta)/(2)-c)/(b))^(2))=(b tan.(theta)/(2))/(1-"tan"^(2)(theta)/(2))`
`rArr ((a^(2)tan.(theta)/(2)-ac))/(b^(2)-(a^(2)"tan"^(2)(theta)/(2)+c^(2)-2ac tan.(theta)/(2)))`
`=(tan.(tehta)/(2))/(1-"tan"^(2)(theta)/(2))`
`rArr (a^(2)tan.(theta)/(2)-ac)(1-"tan"^(2)(theta)/(2))`
`= b^(2) tan.(theta)/(2)-a^(2)"tan"^(3)(theta)/(2)-c^(2)tan.(theta)/(2)+2 ac "tan"^(2)(theta)/(2)`
`= a^(2) tan.(theta)/(2)-ac - a^(2)"tan"^(3)(theta)/(2)+ac "tan"^(2)(theta)/(2)`
`rArr b^(2) tan.(theta)/(2)-a^(2)"tan"^(3)(theta)/(2)-c^(2)tan.(thetA)/(2)+2 ac "tan"^(2)(thetA)/(2)`
`rArr (a^(2)+c^(2)-b^(2))tan.(theta)/(2)=ac[1+ "tan"^(2)(thetA)/(2)]`
`rArr (2 tan.(theta)/(2))/(1+ "Tan"^(2)(thetA)/(2))=(2ac)/(c^(2)+a^(2)-b^(2))`
`rArr sin theta = (2ac)/(c^(2)+a^(2)-b^(2))`
`rArr sin theta = (2ac)/(c^(2)+a^(2)-b^(2))`
Similarly, `sin phi=(2bc)/(a^(2)-b^(2)-c^(2))`