Expert Answers
sciencesolve eNotes educator| Certified Educator

Since the content of the problem is ambiguous and since it is not specified if differentiation is needed, allow for the problem to be solve considering d as a constant.

Factoring out d yields:

`d(a/cos hat A + b/cos hat B + c/cos hat C) = 0 =gt a/cos hat A + b/cos hat B + c/cos hat C = 0`

You should use the following notation `a=BC, b=AC, c=AB` .

You need to remember the law of cosines such that:

`a^2 = b^2 + c^2 - 2ab*cos hat A`

`cos hat A = (b^2 + c^2 - a^2)/(2bc)`

Reasoning by analogy yields:

`cos hat B`  = (a^2 + c^2 - b^2)/(2ac)

`cos hat C ` = (a^2 + b^2 - c^2)/(2ab)

You should substitute `(b^2 + c^2 - a^2)/(2bc)`  for `cos hatA ` , `(a^2 + c^2 - b^2)/(2ac)`  for `cos hat B`  and (a^2 + b^2 - c^2)/(2ab) for `cos hat C ` such that:

`(2abc)/(b^2 + c^2 - a^2) + (2abc)/(a^2 + c^2 - b^2) + (2abc)/(a^2 + b^2 - c^2) = 0`

You may divide by 2abc such that:

`1/(b^2 + c^2 - a^2) + 1/(a^2 + c^2 - b^2) + 1/(a^2 + b^2 - c^2) = 0`

You need to bring the terms to a common denominator such that:

`(a^2 + c^2 - b^2)(a^2 + b^2 - c^2) + (a^2 + b^2 - c^2)(b^2 + c^2 - a^2) + (b^2 + c^2 - a^2)(a^2 + c^2 - b^2) = 0`

You need to open the brackets such that:

`a^4 + a^2b^2 - a^2c^2 + a^2c^2 + b^2c^2 - c^4 - a^2b^2 - b^4 + b^2c^2 + a^2b^2 + a^2c^2 - a^4 + b^4 + b^2c^2 - a^2b^2 - b^2c^2 - c^4 + a^2c^2 + a^2b^2 + b^2c^2 - b^4 + a^2c^2 + c^4 - b^2c^2 - a^4 - a^2c^2 + a^2b^2 = 0`

Reducing like terms yields:

`2b^2c^2 + 2a^2c^2 + 2a^2b^2- c^4 - b^4 - a^4 = 0`

`(-a^2-b^2-c^2)^2 = 0 =gt -a^2-b^2-c^2 = 0 =gt a=b=c=0`

Hence, the identity is verified for a=b=c=0, but this is impossible since a,b,c are the length of the sides of triangle.

jeew-m eNotes educator| Certified Educator

Assume a standard notation for ABC triangle.


Then by law of Sines we can write;


a/SinA = b/SinB = c/SinC = k where k is a constant.



Lets consider the law of sines seperately.
Then ;
a/SinA = k

a = k*sinA


Now let us differntiate this with respect to A.
   da/dA = k*cosA
da/cosA = k*dA--------(1)


Similarly we can write;
db/cosB = k*dB------(2)
dc/cosC = k*dC------(3)

For a triangle A+B+C = pi


By differentiation we get dA+dB+dC = 0

(1)+(2)+(3)
da/cosA+db/cosB+dc/cosC = k(dA+dB+dC)

But we know that dA+dB+dC = 0

Therefore we get
da/cosA+db/cosB+dc/cosC = k(0) = 0

So ;
da/cosA+db/cosB+dc/cosC =0