If A+B+C=pi then find the value of Share with your friends Share 12 Anupam Sharma answered this Dear Student, A+B+c=πsin(A+B+C)sin(B)cos(C)-sin(B)0tan(A)cos(A+B)-tan(A)0As A+B=π-C=sin(π)sin(B)cos(C)-sin(B)0tan(A)cos(π-C)-tan(A)0=0sin(B)cos(C)-sin(B)0tan(A)-cos(C)-tan(A)0Expanding along R1 we get0-sin(B)0+cos(C)tan(A)+cos(C)[0+sin(B)tan(A)]=-sin(B)cos(C)tan(A)+cos(C)sin(B)tan(A)=0 Regards 18 View Full Answer Lokendra Soni answered this given A+B+C= pie. so in the determinant, sin(A+B+C)= sin(pie) = 0. and cos(A+B)= cos(pie - C) = -cos (C). Put these value in determinant and expand along R1 you will get answer=> zero. So the answer is 0(zero) 10
