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Question: It is given that \[{{\Delta }_{1}}=\left| \begin{matrix} x & \sin \theta & \cos \theta \\\ ...

It is given that Δ1=∣xsin⁡θcos⁡θ −sin⁡θ−x1 cos⁡θ1x ∣{{\Delta }_{1}}=\left| \begin{matrix} x & \sin \theta & \cos \theta \\\ -\sin \theta & -x & 1 \\\ \cos \theta & 1 & x \\\ \end{matrix} \right| and Δ2=∣xsin⁡2θcos⁡2θ −sin⁡2θ−x1 cos⁡2θ1x ∣,x≠0;{{\Delta }_{2}}=\left| \begin{matrix} x & \sin 2\theta & \cos 2\theta \\\ -\sin 2\theta & -x & 1 \\\ \cos 2\theta & 1 & x \\\ \end{matrix} \right|,x\ne 0; then for all θ∈(0,π2)\theta \in \left( 0,\dfrac{\pi }{2} \right) :
A. Δ1−Δ2=x(cos⁡2θ−cos⁡4θ) B. Δ1+Δ2=−2x3 C. Δ1−Δ2=−2x3 D. Δ1−Δ2=−2(x3+x−1) \begin{aligned} & \text{A}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=x\left( \cos 2\theta -\cos 4\theta \right) \\\ & \text{B}\text{. }{{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}} \\\ & \text{C}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=-2{{x}^{3}} \\\ & \text{D}\text{. }{{\Delta }_{1}}-{{\Delta }_{2}}=-2\left( {{x}^{3}}+x-1 \right) \\\ \end{aligned}

Explanation

Solution

First we have to solve the determinants given in the question. By simplifying the given determinants we find the values of Δ1 andΔ2{{\Delta }_{1}}\ and {{\Delta }_{2}}. Then, as given in the options we calculate the value of Δ1+Δ2&Δ1−Δ2{{\Delta }_{1}}+{{\Delta }_{2}}\And {{\Delta }_{1}}-{{\Delta }_{2}} and compare the values obtained with the given options to find the correct option.

Complete step-by-step answer :
We have been given Δ1=∣xsin⁡θcos⁡θ −sin⁡θ−x1 cos⁡θ1x ∣{{\Delta }_{1}}=\left| \begin{matrix} x & \sin \theta & \cos \theta \\\ -\sin \theta & -x & 1 \\\ \cos \theta & 1 & x \\\ \end{matrix} \right| and Δ2=∣xsin⁡2θcos⁡2θ −sin⁡2θ−x1 cos⁡2θ1x ∣,x≠0;{{\Delta }_{2}}=\left| \begin{matrix} x & \sin 2\theta & \cos 2\theta \\\ -\sin 2\theta & -x & 1 \\\ \cos 2\theta & 1 & x \\\ \end{matrix} \right|,x\ne 0;
Now, let us first simplify the given determinants to get the values of Δ1&Δ2{{\Delta }_{1}}\And {{\Delta }_{2}}.
Now, we know that if we have determinant of order 3×33\times 3 of the form ∣a1b1c1 a2b2c2 a3b3c3 ∣\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|, then the value of determinant is defined as a1(b2c3−c2b3)−b1(a1c3−c1a3)+c1(a2b3−b2a3){{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{b}_{1}}\left( {{a}_{1}}{{c}_{3}}-{{c}_{1}}{{a}_{3}} \right)+{{c}_{1}}\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right) .
Now, let us first consider Δ1=∣xsin⁡θcos⁡θ −sin⁡θ−x1 cos⁡θ1x ∣{{\Delta }_{1}}=\left| \begin{matrix} x & \sin \theta & \cos \theta \\\ -\sin \theta & -x & 1 \\\ \cos \theta & 1 & x \\\ \end{matrix} \right|
When we compare the given determinant with the general form, the value is defined as Δ1=x(−x×x−1×1)−sin⁡θ(−sin⁡θ×x−cos⁡θ×1)+cos⁡θ(−sin⁡θ×1−(−x×cos⁡θ)){{\Delta }_{1}}=x\left( -x\times x-1\times 1 \right)-\sin \theta \left( -\sin \theta \times x-\cos \theta \times 1 \right)+\cos \theta \left( -\sin \theta \times 1-\left( -x\times \cos \theta \right) \right)
Now, simplifying further we get
Δ1=x(−x2−1)−sin⁡θ(−sin⁡θ×x−cos⁡θ)+cos⁡θ(−sin⁡θ−(−x×cos⁡θ)){{\Delta }_{1}}=x\left( -{{x}^{2}}-1 \right)-\sin \theta \left( -\sin \theta \times x-\cos \theta \right)+\cos \theta \left( -\sin \theta -\left( -x\times \cos \theta \right) \right)
Δ1=x(−x2−1)−sin⁡θ(−xsin⁡θ−cos⁡θ)+cos⁡θ(−sin⁡θ+xcos⁡θ){{\Delta }_{1}}=x\left( -{{x}^{2}}-1 \right)-\sin \theta \left( -x\sin \theta -\cos \theta \right)+\cos \theta \left( -\sin \theta +x\cos \theta \right)
Δ1=−x3−x+xsin⁡2θ+sin⁡θcos⁡θ−sin⁡θcos⁡θ+xcos⁡2θ{{\Delta }_{1}}=-{{x}^{3}}-x+x{{\sin }^{2}}\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{{\cos }^{2}}\theta
Now, simply solve the operations, we get
Δ1=−x3−x+x(sin⁡2θ+cos⁡2θ){{\Delta }_{1}}=-{{x}^{3}}-x+x\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)
Now, we know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
So, substituting the value we get

& {{\Delta }_{1}}=-{{x}^{3}}-x+x\times 1 \\\ & {{\Delta }_{1}}=-{{x}^{3}}.............(i) \\\ \end{aligned}$$ Now, let us consider second determinant $${{\Delta }_{2}}=\left| \begin{matrix} x & \sin 2\theta & \cos 2\theta \\\ -\sin 2\theta & -x & 1 \\\ \cos 2\theta & 1 & x \\\ \end{matrix} \right|$$ When we compare the given determinant with the general form, the value is defined as $${{\Delta }_{2}}=x\left( -x\times x-1\times 1 \right)-\sin 2\theta \left( -\sin 2\theta \times x-\cos 2\theta \times 1 \right)+\cos 2\theta \left( -\sin 2\theta \times 1-\left( -x\times \cos 2\theta \right) \right)$$ Now, simplifying further we get $${{\Delta }_{2}}=x\left( -{{x}^{2}}-1 \right)-\sin 2\theta \left( -x\sin 2\theta -\cos 2\theta \right)+\cos 2\theta \left( -\sin 2\theta +x\cos 2\theta \right)$$ $${{\Delta }_{2}}=-{{x}^{3}}-x+x{{\sin }^{2}}2\theta +\sin 2\theta \cos 2\theta -\sin 2\theta \cos 2\theta +x{{\cos }^{2}}2\theta $$ Now, simply solve the operations, we get $$\begin{aligned} & {{\Delta }_{2}}=-{{x}^{3}}-x+x{{\sin }^{2}}2\theta +x{{\cos }^{2}}2\theta \\\ & {{\Delta }_{2}}=-{{x}^{3}}-x+x\left( {{\sin }^{2}}2\theta +{{\cos }^{2}}2\theta \right) \\\ \end{aligned}$$ Now, we know that $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$ So, substituting the value we get $$\begin{aligned} & {{\Delta }_{2}}=-{{x}^{3}}-x+x\times 1 \\\ & {{\Delta }_{2}}=-{{x}^{3}}.............(ii) \\\ \end{aligned}$$ When we add equation (i) and equation (ii), we get $$\begin{aligned} & {{\Delta }_{1}}+{{\Delta }_{2}}=-{{x}^{3}}+\left( -{{x}^{3}} \right) \\\ & {{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}} \\\ \end{aligned}$$ When we compare the obtained values with the options, we get that the option B is the correct answer. **Note** : We added the equation (i) and equation (ii) first and we get the answer. If we didn’t get any match with the options, we subtract both the equations. $$\begin{aligned} & {{\Delta }_{1}}-{{\Delta }_{2}}=-{{x}^{3}}-\left( -{{x}^{3}} \right) \\\ & {{\Delta }_{1}}-{{\Delta }_{2}}=-{{x}^{3}}+{{x}^{3}} \\\ & {{\Delta }_{1}}-{{\Delta }_{2}}=0 \\\ \end{aligned}$$