1.

किसी `triangleABC` में सिद्ध करे कि `2[asin^(2)""(C)/(2) +csin^(2)""(A)/(2)] = c+a-b`

Answer» बायाँ पक्ष ` = 2[asin^(2)""(C)/(2)+csin^(2)""(A)/(2)]`
` = 2 [a((s-a)(s-b))/(ab) +c((s-b)(s-c))/(bc)]`
` =2[((s-a)(s-b))/(b) +((s-b)(s-c))/(b)]`
` = 2(s-b)[((s-a)+(s-c))/(b)]`
` = 2(s-b)((2s-a-c))/(b) = (2s-2b) *(b)/(b)`
` = (a+b+c-2b) = a +c - b = ` दायाँ पक्ष


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