If \bigtriangleup _{1}=\begin{vmatrix} x &\sin \theta &\cos \theta \\ -\sin \theta &-x &1 \\ \cos \theta & 1 & x \end{vmatrix}  and 

\bigtriangleup _{2}=\begin{vmatrix} x &\sin 2\theta &\cos 2\theta \\ -\sin 2\theta &-x &1 \\ \cos2 \theta & 1 & x \end{vmatrix} , x\neq0 ; then 

for all \theta \epsilon (0,\frac{\pi}{2}) :

  • Option 1)

    \triangle_{1}-\triangle_{2}=-2x^{3}

  • Option 2)

    \triangle_{1}-\triangle_{2}=x(cos2\theta-cos4\theta)

  • Option 3)

    \triangle_{1}+\triangle_{2}=-2(x^{3}+x-1)

  • Option 4)

    \triangle_{1}+\triangle_{2}=-2x^{3}

Answers (1)

\Delta _{1}=x(-x^{2}-1)-sin\theta(-xsin\theta-cos\theta)+cos\theta(-sin\theta+xcos\theta)

       =-x^{3}-x+xsin^{2}\theta+sin\theta cos\theta-cos\theta sin\theta+xcos^{2}\theta

      =-x^{3}-x+x(sin^{2}\theta+cos^{2}\theta)

      =-x^{3}

\bigtriangleup _{2}\: \: is \: \: also\: \: -x^{3}

So, \bigtriangleup _{1}+\bigtriangleup _{2}=-2x^{3}

So, option (4) is correct.


Option 1)

\triangle_{1}-\triangle_{2}=-2x^{3}

Option 2)

\triangle_{1}-\triangle_{2}=x(cos2\theta-cos4\theta)

Option 3)

\triangle_{1}+\triangle_{2}=-2(x^{3}+x-1)

Option 4)

\triangle_{1}+\triangle_{2}=-2x^{3}

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