Trắc nghiệm Giá trị có lượng giác của một cung có đáp án (Thông hiểu)

A. $cot\alpha \tan \alpha =1,\alpha \ne \frac{k\pi }{2},k\in Z$

B. $1+{\tan }^2\alpha =\frac{1}{co{s}^2\alpha },\alpha \ne k\pi ,k\in Z$

C. ${\sin }^2\alpha +{\cos }^2\beta =1$

D. $1+co{t}^2\alpha =\frac{1}{{\sin }^2\alpha },\alpha \ne \frac{\pi }{2}+k\pi ,k\in Z$

A. ${\sin }^2\alpha +{\cos }^2\alpha =1$

B. $1+\mathrm{ta}{n}^2\alpha =\frac{1}{co{s}^2\alpha }(\alpha \ne \frac{\pi }{2}+k\pi ,k\in Z)$

C. $1+co{t}^2\alpha =\frac{1}{{\sin }^2\alpha },(\alpha \ne k\pi ,k\in Z)$

D. $\tan \alpha +cot\alpha =1,(\alpha \ne \frac{k\pi }{2},k\in Z)$

A. $-\frac{15}{3};\frac{2}{3}$

B. $\frac{2}{3};\frac{-5}{12}$

C. $-\frac{5}{13};\frac{5}{12}$

D. $\frac{5}{13};\frac{-5}{12}$

A. $cot\alpha =2$

B. $cot\alpha =\frac{1}{4}$

C. $cot\alpha =\frac{1}{2}$

D. $cot\alpha =\sqrt{2}$

A. $\frac{1}{2}$

B. $-\frac{1}{2}$

C. 1

D. 3

A. ${(\sin x+\cos x)}^2=1+2\sin x\cos x$

B. ${(\sin x-\cos x)}^2=1-2\sin x\cos x$

C. ${\sin }^{4}x+{\cos }^{4}x=1-2{\sin }^{2}x{\cos }^{2}x$

D. ${\sin }^{6}x+{\cos }^{6}x=1+3{\sin }^{2}x{\cos }^{2}x$

A. $\sin \left(\frac{\pi }{2}+\alpha \right)>0$

B. $\sin \left(\frac{\pi }{2}+\alpha \right)\ge 0$

C. $\sin \left(\frac{\pi }{2}+\alpha \right)<0$

D. $\sin \left(\frac{\pi }{2}+\alpha \right)\le 0$

A. tana > 0, cota > 0

B. tana < 0, cota < 0

C. tana > 0, cota < 0

D. tana < 0, cota > 0

A. $\tan \alpha =\frac{-\sqrt{2}}{4}$

B. $\tan \alpha =-2\sqrt{2}$

C. $\tan \alpha =2\sqrt{2}$

D. $\tan \alpha =\frac{\sqrt{2}}{4}$

A. $\cot \alpha =\frac{2\sqrt{5}}{5}$

B. $\cot \alpha =2\sqrt{5}$

C. $\cot \alpha =-2\sqrt{5}$

D. $\cot \alpha =\frac{-2\sqrt{5}}{5}$

A. $\frac{5-\sqrt{7}}{4}hay\frac{5+\sqrt{7}}{4}$

B. $\frac{5-\sqrt{5}}{7}hay\frac{5+\sqrt{5}}{7}$

C. $\frac{2-\sqrt{3}}{5}hay\frac{2+\sqrt{3}}{5}$

D. $\frac{3-\sqrt{2}}{5}hay\frac{3+\sqrt{2}}{5}$

A. 0

B. 1

C. 2

D. 4

A. 0

B. 1

C. 2

D. -1

A. $\sin {225}^0=\frac{-\sqrt{2}}{2}$

B. $cos{225}^0=\frac{-\sqrt{2}}{2}$

C. $\tan {225}^0=-1$

D. $\cot {225}^0=1$

A. $-3-\sqrt{3}$

B. $2-3\sqrt{3}$

C. $\frac{2\sqrt{3}}{\sqrt{3}-1}$

D. $\frac{1-\sqrt{3}}{\sqrt{3}}$