Solving,
⇒[2 sin 30°-cot 45°-2 cos 60°-sin 90°][tan 45°cosec 30°sec 60°cos 0°]⇒[2×21−1−2×21−1][1221]⇒[1−1−1−1][1221]⇒[1×1+(−1)×2(−1)×1+(−1)×21×2+(−1)×1(−1)×2+(−1)×1]⇒[1−2−1−22−1−2−1]⇒[−1−31−3].
Hence, [2 sin 30°-cot 45°-2 cos 60°-sin 90°][tan 45°cosec 30°sec 60°cos 0°]=[−1−31−3].