JEE Main 201910 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 10 April
- Shift
- Morning Shift
- Subject
- Mathematics
- Chapter
- Determinants
JEE Main 2019 · 10 April · Morning Shift · Mathematics · Determinants
If Δ _ 1 = | cases x & sinθ & cosθ - sinθ & - x & 1 cosθ & 1 & x...
If {\Delta}_{1} = \left|\begin{cases}x & sin\theta & cos\theta \\ - sin\theta & - x & 1 \\ cos\theta & 1 & x\end{cases}\right| and {\Delta}_{2} = \left|\begin{cases}x & \sin 2\theta & \cos 2\theta \\ - \sin 2\theta & - x & 1 \\ \cos 2\theta & 1 & x\end{cases}\right|, then for all :
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
{\Delta}_{1} = \left|\begin{cases}x & sin\theta \\ - sin\theta & - x \\ cos\theta & 1\end{cases}\right|= x\left(- {x}^{2} - 1\right) - sin\theta\left(- xsin\theta - cos\theta\right) + cos\theta ( - sin\theta + xcos\theta )= - {x}^{3} - x + xsi{n}^{2}\theta + sin\theta cos \theta - sin\theta cos\theta + xco{s}^{2}\theta= - {x}^{3} - x + x\left(\sin^{2} \theta + \cos^{2} \theta\right)= - {x}^{3}{\Delta}_{2} = - {x}^{3}∴ {\Delta}_{1} + {\Delta}_{2} = - 2{x}^{3}$