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