Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# ABC is a triangle, then find ${a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)$.A. 0B. 1C. ${a^2} + {b^2} + {c^2}$D. $2\left( {{a^2} + {b^2} + {c^2}} \right)$

Last updated date: 13th Jun 2024
Total views: 51.6k
Views today: 1.51k
Verified
51.6k+ views
Hint: First we will convert all $\cos$ function into $\sin$ function using trigonometry identity. Then using the sine law, we will find the value of $\sin A$, $\sin B$ and $\sin C$. After that we will substitute the value of $\sin A$, $\sin B$ and $\sin C$ in the given expression.

Formula used:
Sine law
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$
Trigonometry identity
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step by step solution:
Given expression is ${a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)$
Now applying the trigonometry identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$= {a^2}\left( {1 - {{\sin }^2}B - 1 + {{\sin }^2}C} \right) + {b^2}\left( {1 - {{\sin }^2}C - 1 + {{\sin }^2}A} \right) + {c^2}\left( {1 - {{\sin }^2}A - 1 + {{\sin }^2}B} \right)$
$= {a^2}\left( {{{\sin }^2}C - {{\sin }^2}B} \right) + {b^2}\left( {{{\sin }^2}A - {{\sin }^2}C} \right) + {c^2}\left( {{{\sin }^2}B - {{\sin }^2}A} \right)$ …..(i)
We know that,
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {{\rm{say}}} \right)$
$\sin A = ak$, $\sin B = bk$, and $\sin C = ck$
Substitute $\sin A = ak$, $\sin B = bk$, and $\sin C = ck$ in the expression (i)
$= {a^2}\left( {{c^2}{k^2} - {b^2}{k^2}} \right) + {b^2}\left( {{a^2}{k^2} - {c^2}{k^2}} \right) + {c^2}\left( {{b^2}{k^2} - {a^2}{k^2}} \right)$
Simplify the above expression
$= {a^2}{c^2}{k^2} - {a^2}{b^2}{k^2} + {b^2}{a^2}{k^2} - {b^2}{c^2}{k^2} + {c^2}{b^2}{k^2} - {c^2}{a^2}{k^2}$
Cancel out the opposite term
$= 0$
Hence option A is the correct option.

Note: Students often make a common mistake to solve the given question. They apply cosine law to solve the given question. But the given question should be solved by using trigonometry identity and the sine law.