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in a triangle ABC, prove that a^2cotA + b^2cotb + c^2cotc = abc\R
question is from the chapter properties of triangles

meghana p , 5 Years ago
Grade
anser 1 Answers
Aditya Gupta

Last Activity: 5 Years ago

note that sinA/a= sinB/b= sinC/c= 1/2R. 
so a= 2RsinA, b= 2RsinB and c= 2RsinC
hence, a^2cotA + b^2cotb + c^2cotc= 4R^2(sin^2A*cosA/sinA + sin^2B*cosB/sinB + sin^2C*cosC/sinC)
= 2R^2(sin2A + sin2B + sin2C)
but by conditional identities we know that sin2A + sin2B + sin2C= 4sinAsinBsinC
so, RHS= 8R^2sinAsinBsinC= 8R^2*(a/2R)*(b/2R)*(c/2R)
= abc/R
kindly approve :))

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