Question

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

Answer 1

Hint: In the above question, we will 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 have been given to prove \[{{a}^{2}}\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)+{{b}^{2}}\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)+{{c}^{2}}\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)=0\]
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:
\[{{a}^{2}}\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)+{{b}^{2}}\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)+{{c}^{2}}\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)\]
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:
\[={{a}^{2}}\left( 1-{{\sin }^{2}}B-1+{{\sin }^{2}}C \right)+{{b}^{2}}\left( 1-{{\sin }^{2}}C-1+{{\sin }^{2}}A \right)+{{c}^{2}}\left( 1-{{\sin }^{2}}A-1+{{\sin }^{2}}B \right)\]
Simplifying the above expression, we get,
\[={{a}^{2}}\left( {{\sin }^{2}}C-{{\sin }^{2}}B \right)+{{b}^{2}}\left( {{\sin }^{2}}A-{{\sin }^{2}}C \right)+{{c}^{2}}\left( {{\sin }^{2}}B-{{\sin }^{2}}A \right)\]
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:
\[={{a}^{2}}\left( {{c}^{2}}{{k}^{2}}-{{b}^{2}}{{k}^{2}} \right)+{{b}^{2}}\left( {{a}^{2}}{{k}^{2}}-{{k}^{2}}{{c}^{2}} \right)+{{c}^{2}}\left( {{k}^{2}}{{b}^{2}}-{{k}^{2}}{{a}^{2}} \right)\]
Taking \[{{k}^{2}}\] as common from the given expression and expanding the bracket, we get,
\[={{k}^{2}}\left[ {{a}^{2}}{{c}^{2}}-{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{a}^{2}}-{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}-{{a}^{2}}{{c}^{2}} \right]\]
On simplifying the above expression, we get,
\[={{k}^{2}}\times 0\]
= 0
Hence, it is proved that
\[{{a}^{2}}\left( {{\cos }^{2}}B-{{\cos }^{2}}C \right)+{{b}^{2}}\left( {{\cos }^{2}}C-{{\cos }^{2}}A \right)+{{c}^{2}}\left( {{\cos }^{2}}A-{{\cos }^{2}}B \right)=0\]

Note: Just remember the sine rule, cosine rule, and other trigonometric identities as it will help you a lot in these types of questions. Also, we can prove the given expression by using the cosine rule.