Đáp án:
$a)
K=-2\sin x\\
b)
L=3\sin x$
Giải thích các bước giải:
$a)
K=\sin(\pi+x)-\cos\left ( \dfrac{\pi}{2}-x \right )+\tan\left ( \dfrac{3\pi}{2}-x \right )+\cot\left ( 2\pi-x \right )\\
=-\sin x-\sin x+\cot x-\cot x\\
=-2\sin x\\
b)
L=\cot(x-2\pi).\cos\left ( x-\dfrac{3\pi}{2} \right )+\cos(x-2\pi)-3\sin(x-\pi)\\
=-\cot(2\pi-x)\cos\left ( \dfrac{3\pi}{2}-x \right )+\cos(2\pi-x)+3\sin(\pi-x)\\
=\cot x.\sin x-\cos x+3\sin x\\
=\dfrac{\cos x}{\sin x}.\sin x-\cos x+3\sin x\\
=\cos x-\cos x+3\sin x\\
=3\sin x$