Respuesta :

gmany

Answer:

cos90° = 0°, tan0° = 0°,

cos(-90°) = 0°, cot270° = 0°

Step-by-step explanation:

[tex]k\in\mathbb{Z}\\\\\sin x=0\iff x=180^ok\\\\\cos x=0\iff x=90^o+180^ok\\\\\tan x=0\iff x=180^ok\\\\\cot x=0\iff x=90^o+180^ok\\\\\csc x\neq0\ \text{for}\ x\in\mathbb{R}-\{180^ok\}\\========================\\\\\sin270^o\neq0\\\\\cos90^o=\cos(90^o+180^o\cdot0)=0\\\\\tan0^o=\tan(180^o\cdot0)=0\\\\\csc(-180^o)\neq0\\\\\cos(-90^o)=\cos(90^o-180^o)=\cos(90^o+180^o\cdot(-1))=0\\\\\cot270^o=\cot(90^o+180^o)=0[/tex]

27andriannakrause

Answer:

cos90° = 0°, tan0° = 0°,cos(-90°) = 0°, cot270° = 0°

Step-by-step explanation:

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