किसी त्रिभुज ABC में सिद्ध कीजिए कि
`(a-b)/(c ) = ( tan. ( A)/(2)-tan. (B)/(2))/(tan. (A)/(2)+tan. (B)/(2))`

Explanation

बायाँ पक्ष ` = (a-b)/(c) = (ksinA-ksinB)/(ksinC) = (sinA-sinB)/(sinC)`
` = (2cos""(A+B)/(2)*sin""(A-B)/(2))/(2sin""(C)/(2)cos""(C)/(2)) = (2sin""(C)/(2)cos""(A-B)/(2))/(2sin""(C)/(2)cos""(C)/(2))`
` = (sin""(A-B)/(2))/(cos""(C)/(2)) = (sin""(A-B)/(2))/(sin((A+B)/(2))) = (sin((A)/(2)-(B)/(2)))/(sin((A)/(2)+(B)/(2)))`
` = (sin""(A)/(2)*cos""(B)/(2)-cos""(A)/(2)*sin""(B)/(2))/(sin""(A)/(2)*cos""(B)/(2)+cos""(A)/(2)*sin""(B)/(2))`
` = ((sin""(A)/(2)*cos""(B)/(2))/(cos""(A)/(2)cos""(B)/(2))-(cos""(A)/(2)sin""(B)/(2))/(cos""(A)/(2)sin""(B)/(2)))/((sin""(A)/(2)cos""(B)/(2))/(cos""(A)/(2)*cos""(B)/(2))+(cos""(A)/(2)*sin""(B)/(2))/(cos""(A)/(2)*cos""(B)/(2)))`[अंश तथा हर को `cos""(A)/(2)*cos""(B)/(2)` से भाग देने पर ]
`= (tan""(A)/(2)-tan""(B)/(2))/(tan""(A)/(2)+tan""(B)/(2)) =` दायाँ पक्ष