In \[\Delta ABC\], find \[{b^2}\cos 2A - {a^2}\cos 2B\].
A. \[{b^2} - {a^2}\]
B. \[{b^2} - {c^2}\]
C. \[{c^2} - {a^2}\]
D. \[{a^2} + {b^2} + {c^2}\]

Hint: We will apply the double angle formula of trigonometry to convert cos into sin. Then apply sine law to simplify the given expression.

Formula used:
Double angle formula
\[\cos 2\theta = 1 - 2{\sin ^2}\theta \]
Sine law:
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

complete step by step solution:
Given expression is
\[{b^2}\cos 2A - {a^2}\cos 2B\]
Applying the double angle formula \[\cos 2\theta = 1 - 2{\sin ^2}\theta \]
\[ = {b^2}\left( {1 - 2{{\sin }^2}A} \right) - {a^2}\left( {1 - 2{{\sin }^2}B} \right)\]
Simplify the above equation
\[ = {b^2} - 2{b^2}{\sin ^2}A - {a^2} + 2{a^2}{\sin ^2}B\]
\[ = {b^2} - {a^2} - 2{b^2}{\sin ^2}A + 2{a^2}{\sin ^2}B\] …..(i)
We know that sine law state that,
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
\[ \Rightarrow \dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b}\]
Cross multiply
\[ \Rightarrow b\sin A = a\sin B\]
Squaring both sides
\[ \Rightarrow {b^2}{\sin ^2}A = {a^2}{\sin ^2}B\]
\[ \Rightarrow {a^2}{\sin ^2}B - {b^2}{\sin ^2}A = 0\]
Substitute \[{a^2}{\sin ^2}B - {b^2}{\sin ^2}A = 0\] in (i)
\[ = {b^2} - {a^2} + 2 \cdot 0\]
\[ = {b^2} - {a^2}\]
Hence option A is the correct option.

Addition information:
If \[\Delta ABC\] is a triangle and AB = c, BC = a , AC = b, it follows some rules that are Cosine law and Sine rule.
The cosine law are
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
The law of sine is
\[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
Half angle formulas are
\[\cos \dfrac{A}{2} = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \]
\[\cos \dfrac{B}{2} = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ca}}} \]
\[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \]
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]
Where s is semi perimeter and \[s = \dfrac{{a + b + c}}{2}\].


Note: Students often make the mistake to apply the double angle formula of cos. They use \[\cos 2\theta = 2{\sin ^2}\theta - 1\] which is incorrect. The correct formula is \[\cos 2\theta = 1 - 2{\sin ^2}\theta \].