Question

If $\Delta_{1} = \begin{vmatrix}x&\sin\theta&\cos\theta\\ -\sin\theta&-x&1\\ \cos\theta&1&x\end{vmatrix} $ and $ \Delta_{2} = \begin{vmatrix}x&\sin2\theta&\cos2 \theta\\ -\sin2\theta&-x&1\\ \cos2\theta& 1&x\end{vmatrix} , x \ne0 ; $ then for all $ \theta \in \left(0, \frac{\pi}{2}\right) :$

• A. $\Delta_1 - \Delta_2 = x (\cos \; 2 \theta - \cos 4 \theta)$
• B. $\Delta_1 + \Delta _2 = - 2x^3$
• C. $\Delta_1 - \Delta _2 = - 2x^3$
• D. $\Delta_1 + \Delta _2 = - 2 (x^3 + x - 1)$

Answer 1

Answer: (B)

$\Delta_{1} =f\left(\theta\right) = \begin{vmatrix}x&\sin\theta&\cos\theta\\ -\sin\theta&-x&1\\ \cos\theta&1&x\end{vmatrix} =-x^{3} $
and $ \Delta_{2} =f\left(2\theta\right) = \begin{vmatrix}x &\sin2\theta&\cos2\theta\\ -\sin2\theta&-x&1\\ \cos2\theta&1&x\end{vmatrix} = -x^{3}$
So $ \Delta_{1} + \Delta_{2} = - 2x^{3}$