Question

If \[C = 2\cos \theta \], then the value of the determinant\[\vartriangle = \left( {\begin{array}{*{20}{c}}
  C&1&0 \\
  1&C&1 \\
  6&1&C
\end{array}} \right)\]is:
(A) \[\dfrac{{\sin 4\theta }}{{\sin \theta }}\]
(B) \[\dfrac{{2{{\sin }^2}\theta }}{{\sin \theta }}\]
(C) \[4{\cos ^2}\theta (2\cos \theta - 1)\]
(D) None of these above

Answer 1

Hint: First do the determinant of \[\vartriangle \]by using this formula:
\[\vartriangle = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)\]
\[dit\left| \vartriangle \right| = {a_{11}}({a_{22}}{a_{33}} - {a_{32}}{a_{23}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{31}}{a_{22}})\]
Then after finding the determinant, put the value of \[C = 2\cos \theta \] and solve the equation.

Complete step-by-step answer:
\[\vartriangle = \left( {\begin{array}{*{20}{c}}
  C&1&0 \\
  1&C&1 \\
  6&1&C
\end{array}} \right)\]
Now do the determinant by using formula \[dit\left| \vartriangle \right| = {a_{11}}({a_{22}}{a_{33}} - {a_{32}}{a_{23}}) - {a_{12}}({a_{21}}{a_{33}} - {a_{23}}{a_{31}}) + {a_{13}}({a_{21}}{a_{32}} - {a_{31}}{a_{22}})\].
Substituting all the terms in the formula given above

\[dit\left| \vartriangle \right| = C({C^2} - 1) - 1(C - 6) + 0(1 - 6C)\]
Multiplying the terms within the brackets
           \[ = {C^3} - C - C + 6\]
Collecting the same values
           \[ = {C^3} - 2C + 6\]
Now put the value of\[C = 2\cos \theta \]in above equation,
             \[ = 8{\cos ^3}\theta - 4\cos \theta + 6\]
Take \[4\cos \theta \] common from the first two terms.
             \[ = 4\cos \theta ({\cos ^2}\theta - 1) + 6\]
In the above equation we use the formula\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \].
\[ = 4\cos \theta ( - {\sin ^2}\theta ) + 6\]
Multiplying the terms within the bracket we get,
            \[ = - 4\cos \theta {\sin ^2}\theta + 6\]
Now no answer match with option but for confirmation takes \[\theta = \dfrac{\pi }{2}\]
First of all we have to put in equation that we get\[\theta = \dfrac{\pi }{2}\] after that we have to match the answer with all option by putting the value \[\theta = \dfrac{\pi }{2}\]
\[dit{\left| \vartriangle \right|_{\theta = \dfrac{\pi }{2}}} = - 4\cos \theta {\sin ^2}\theta + 6\]
                 \[ = - 4\cos \left( {\dfrac{\pi }{2}} \right){\sin ^2}\left( {\dfrac{\pi }{2}} \right) + 6\]
Put the value of \[\cos \left( {\dfrac{\pi }{2}} \right) = 0\]and \[\sin \dfrac{\pi }{2} = 1\]
                 \[ = 6\]
So, we get answer 6 now put the \[\theta = \dfrac{\pi }{2}\] and check for option A, option B and option C.
\[{\left( {\dfrac{{\sin 4\theta }}{{\sin \theta }}} \right)_{\theta = \dfrac{\pi }{2}}} = \dfrac{{\sin 4\left( {\dfrac{\pi }{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2}} \right)}}\]
Put the value of \[\sin \pi = 0\]and \[\sin \dfrac{\pi }{2} = 1\]
\[ = 0\]
\[{\left( {\dfrac{{2{{\sin }^2}\theta }}{{\sin \theta }}} \right)_{\theta = \dfrac{\pi }{2}}} = \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{2}} \right)}}{{\sin \left( {\dfrac{\pi }{2}} \right)}}\]
\[ = 2\]
\[{\left[ {4{{\cos }^2}\theta (2\cos \theta - 1)} \right]_{\theta = \dfrac{\pi }{2}}} = 4{\cos ^2}\left( {\dfrac{\pi }{2}} \right)\left( {2\cos \left( {\dfrac{\pi }{2}} \right) - 1} \right)\]
Put the value of \[\cos \dfrac{\pi }{2} = 0\].
\[ = 0\]

For \[\theta = \dfrac{\pi }{2}\] does not satisfy any option, so the answer is option D is correct. i.e. None of these.

Note:
If there is no answer match and there is given option none of these then try this by taking different values of \[\theta \] and match the answer with option. Students are advised to use trigonometric identities to make the solution easier. Also while calculating the value of determinant students make the mistake of multiplying negative one to the middle term but they should check if the formula already has minus sign along with the middle term and not repeat it because then the value of determinant will be wrong.