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Mathematics

Evaluate without using tables :

[2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \ \sec 60^\circ & \sin 90^\circ \end{bmatrix}

Matrices

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Answer

Solving,

⇒[2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]⇒[2×12−2×12−11]×[1221]⇒[1−1−11]×[1221]⇒[(1)(1)+(−1)(2)(1)(2)+(−1)(1)(−1)(1)+(1)(2)(−1)(2)+(1)(1)]⇒[1−22−1−1+2−2+1]⇒[−111−1].\Rightarrow \begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \ \sec 60^\circ & \sin 90^\circ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times \dfrac{1}{2} & -2 \times \dfrac{1}{2} \ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} (1)(1) + (-1)(2) & (1)(2) + (-1)(1) \ (-1)(1) + (1)(2) & (-1)(2) + (1)(1) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 - 2 & 2 - 1 \ -1 + 2 & -2 + 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}.

Hence, [2cos⁡60∘−2sin⁡30∘−tan⁡45∘cos⁡0∘]×[cot⁡45∘cosec⁡30∘sec⁡60∘sin⁡90∘]=[−111−1]\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \ -\tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \cosec 30^\circ \ \sec 60^\circ & \sin 90^\circ \end{bmatrix} = \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}

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