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If A+B+C=180 prove that cotA.cotB.cotC=cotA+cotB+cotC-cosecA.cosecB.cosecC.

If A+B+C=180 prove that cotA.cotB.cotC=cotA+cotB+cotC-cosecA.cosecB.cosecC.

Grade:11

1 Answers

Aarushi Ahlawat
41 Points
5 years ago
First use the given condition A+B+C=180. start with cos of both the side. after expanding the cos and sin using sum of angles identity, divide both the side by sin(A)sin(B)sin(C) and you will get the desired relation that need to be proved. Here it is how to do it:
cos(A+B+C) = cos(180)=-1
=cos(A+B)cos(C)-sin(A+B)sinC = -1
=\left (cos(A)cos(B)-sin(A)sin(B) \right )cos(C)-\left ( sin(A)cos(B)+cos(A)sin(B) \right )sin(C) = -1
=cos(A)cos(B)cos(C)-sin(A)sin(B)cos(C)-sin(A)cos(B)sin(C)-cos(A)sin(B)sin(C) = -1
 
=\frac{cos(A)cos(B)cos(C)-sin(A)sin(B)cos(C)-sin(A)cos(B)sin(C)-cos(A)sin(B)sin(C)}{sin(A)sin(B)sin(C)} =\frac{-1}{sin(A)sin(B)sin(C)}
=cot(A)cot(B)cot(C)-cot(C)-cot(B)-cot(A)=-cosec(A)cosec(B)cosec(C)
=cot(A)cot(B)cot(C)=cot(A)+cot(B)+cot(C)-cosec(A)cosec(B)cosec(C)

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