Question

In any\[\Delta ABC\] , the simplified form of \[\left( {cos{\text{ }}\dfrac{{2A}}{{{a^2}}}} \right) - \left( {cos{\text{ }}\dfrac{{2B}}{{{b^2}}}} \right)\] is
A. \[{a^2} - {b^2}\]
B. \[\dfrac{1}{{\left( {{a^2} - {b^2}} \right)}}\]
C. $\dfrac{1}{{{a^2}}} - \dfrac{1}{{{b^2}}}$
D. ${a^2} + {b^2}$

Answer 1

Hint: Let us take that A, B and C is angle of triangle \[\Delta ABC\] and also that the formula for \[cos2\theta \] will be \[cos2\theta = 1-2si{n^2}\theta \]. Now further we should know the three rules of sine one of which we will be using to further solve.

Formula Used:
We know by the formula of \[cos2\theta \] that
\[cos2\theta = 1-2si{n^2}\theta \]

Sine rule:
\[\dfrac{{a{\text{ }}}}{{sinA}} = \dfrac{b}{{sinB}} = \dfrac{c}{{sinC}} = k\]

Complete step by step Solution:
Given
According to the question given
\[ = \left( {cos{\text{ }}\dfrac{{2A}}{{{a^2}}}} \right) - \left( {cos{\text{ }}\dfrac{{2B}}{{{b^2}}}} \right)\]

\[ = \dfrac{{\left( {1-2si{n^2}A} \right)}}{{{a^2}}} - \dfrac{{\left( {1-2si{n^2}B} \right)}}{{{b^2}}}\]

By further solving the above equation
$ = = \dfrac{1}{{{a^2}}}-\dfrac{1}{{{b^2}}} + 2\left( {\dfrac{{\left( {si{n^2}B} \right)}}{{{a^2}}} - \dfrac{{\left( {si{n^2}A} \right)}}{{{b^2}}}} \right)$
Now according to sine rule used in trigonometry
\[\dfrac{{a{\text{ }}}}{{sinA}} = \dfrac{b}{{sinB}} = \dfrac{c}{{sinC}} = k\]

\[ = \dfrac{1}{{{a^2}}}-\dfrac{{1{\text{ }}}}{{2{k^2}}}-\dfrac{1}{{{b^2}}} + \dfrac{1}{{2{k^2}}}\]

Thus, the value of $k = 0$, and by further solving the equation we get

= \[\dfrac{1}{{{a^2}}}-\dfrac{1}{{{b^2}}}\]

Hence, the correct option is C.

Note: Before solving any type of trigonometry question knowing the formulas can help a lot in solving. Also with that assessing and understanding the questions is also a necessity as which formula to use is a key component. As in this question, we used \[cos2\theta \] and the sine rule to solve.