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Q.Simplify - cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cos\theta \begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix} + \sin\theta \begin{bmatrix}\sin\theta & -\cos\theta \\ \cos\theta & \sin\theta\end{bmatrix}

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2022Subjective· 2mImportance★★★★★
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Distribute the scalars into each matrix, add, and use cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1.

cos⁡θ[cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡2θsin⁡θcos⁡θ−sin⁡θcos⁡θcos⁡2θ]\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix} = \begin{bmatrix}\cos^2\theta & \sin\theta\cos\theta\\-\sin\theta\cos\theta & \cos^2\theta\end{bmatrix}

sin⁡θ[sin⁡θ−cos⁡θcos⁡θsin⁡θ]=[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix} = \begin{bmatrix}\sin^2\theta & -\sin\theta\cos\theta\\\sin\theta\cos\theta & \sin^2\theta\end{bmatrix}

Adding the two matrices entrywise: …

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