Answer
Verified
106.8k+ 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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main