Question

 In any \[\Delta ABC\], prove that
\[\dfrac{\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)}{b+c}+\dfrac{\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)}{c+a}+\dfrac{\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)}{a+b}=0\]

Answer 1

Hint: In the above question, we will just put the values of \[{{\cos }^{2}}B\], \[{{\cos }^{2}}C\]and \[{{\cos }^{2}}A\] in terms of \[{{\sin }^{2}}B\], \[{{\sin }^{2}}C\] and \[{{\sin }^{2}}A\] respectively and then we will use the sine rule of the triangle. The trigonometric identities that we will use in the above question are as follows:
\[{{\cos }^{2}}A=1-{{\sin }^{2}}A\]
\[{{\cos }^{2}}B=1-{{\sin }^{2}}B\]
\[{{\cos }^{2}}C=1-{{\sin }^{2}}C\]

Complete step by step answer:
We know that the sine rule of the triangle is as follows:
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
Let the ratio is equal to ‘k’, we get,
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k\]
\[\Rightarrow \sin A=ak,\text{ }\sin B=bk\text{ and }\sin C=ck\]
Taking the left-hand side of the given expression which we have to prove is as follows:
\[\dfrac{\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)}{b+c}+\dfrac{\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)}{c+a}+\dfrac{\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)}{a+b}=0\]
Substituting the values of \[{{\cos }^{2}}B\], \[{{\cos }^{2}}C\]and \[{{\cos }^{2}}A\] in terms of \[{{\sin }^{2}}B\], \[{{\sin }^{2}}C\] and \[{{\sin }^{2}}A\] respectively we get the expression as follows:
\[\dfrac{1-{{\sin }^{2}}B-1+{{\sin }^{2}}C}{b+c}+\dfrac{1-{{\sin }^{2}}C-1+{{\sin }^{2}}A}{c+a}+\dfrac{1-{{\sin }^{2}}A-1+{{\sin }^{2}}B}{a+b}\]
Simplifying the above expression, we get,
\[=\dfrac{{{\sin }^{2}}C-{{\sin }^{2}}B}{b+c}+\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}C}{c+a}+\dfrac{{{\sin }^{2}}B-{{\sin }^{2}}A}{a+b}\]
On substituting the values of sin A, sin B, and sin C from the sine rule of the triangle we get the expression as follows:
\[=\dfrac{{{k}^{2}}{{c}^{2}}-{{k}^{2}}{{b}^{2}}}{b+c}+\dfrac{{{k}^{2}}{{a}^{2}}-{{k}^{2}}{{c}^{2}}}{c+a}+\dfrac{{{k}^{2}}{{b}^{2}}-{{k}^{2}}{{a}^{2}}}{a+b}\]
Taking \[{{k}^{2}}\] as common from the given expression, we get,
\[={{k}^{2}}\left[ \dfrac{\left( {{c}^{2}}-{{b}^{2}} \right)}{b+c}+\dfrac{\left( {{a}^{2}}-{{c}^{2}} \right)}{c+a}+\dfrac{{{b}^{2}}-{{a}^{2}}}{a+b} \right]\]
On simplifying the above expression, we get,
\[={{k}^{2}}\left[ c-b+a-c+b-a \right]\]
\[={{k}^{2}}\times 0\]
= 0
Hence, it is proved that
\[\dfrac{\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)}{b+c}+\dfrac{\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)}{c+a}+\dfrac{\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)}{a+b}=0\]

Note: We can also prove the given expression by using cosine rule but it will be a little lengthy calculation. We must be very careful as the method uses many trigonometric formulas and any small mistake, be it a change in sign, will affect the final answer. It will also lead to wastage of time in the exams.