10 câu Trắc nghiệm Ôn tập Toán 10 Chương 6 có đáp án (Thông hiểu)

Đáp án A

$$\begin{gathered} C=\frac{1}{\sin {10}^0}+\frac{\sqrt{3}}{\cos {10}^0}=\frac{\cos {10}^0+\sqrt{3}\sin {10}^0}{\sin {10}^0\cos {10}^0} \\ =\frac{\frac{1}{2}\cos {10}^0+\frac{\sqrt{3}}{2}\sin {10}^0}{\frac{2\sin {10}^0\cos {10}^0}{4}}\frac{4\sin {40}^0}{\sin {20}^0}=8\cos {20}^0 \end{gathered}$$

Đáp án C

$\cot x=\frac{3}{4}\Rightarrow \tan x=\frac{4}{3}$ Phương án A sai

$$\begin{gathered} 1+{\cot }^{2}x=\frac{1}{{\sin }^{2}x}\iff 1+{\left(\frac{3}{4}\right)}^2=\frac{1}{{\sin }^{2}x} \\ \iff {\sin }^{2}x=\frac{16}{25}\iff \mathrm{s}\text{inx}=\pm \frac{4}{5} \end{gathered}$$

Mà ${90}^0<x<{180}^0\Rightarrow \mathrm{s}\text{inx}=\frac{4}{5}$ chọn C

Vì ${\sin }^{2}x=\frac{16}{25}\Rightarrow {\cos }^{2}x=\frac{9}{25}\iff \cos x=\pm \frac{3}{5}$

Mà ${90}^0<x<{180}^0\Rightarrow \cos \text{x}=-\frac{3}{5}$ phương án B sai

Đáp án D

$$\begin{gathered} \cos \left(-\frac{23\pi }{6}\right)-\frac{1}{{\cos }^2\frac{16\pi }{3}}+\cot \frac{23\pi }{6} \\ =\cos \left(-4\pi +\frac{\pi }{6}\right)-\frac{1}{\frac{1+\cos \frac{32\pi }{3}}{2}}+\cot \left(4\pi -\frac{\pi }{6}\right) \\ =\cos \frac{\pi }{6}-\frac{2}{1+\cos \left(10\pi +\frac{2\pi }{3}\right)}+\cot \left(-\frac{\pi }{6}\right) \\ =\cos \frac{\pi }{6}-\frac{2}{1+\cos \frac{2\pi }{3}}-\cot \frac{\pi }{6} \\ =\frac{\sqrt{3}}{2}-\frac{2}{1-\frac{1}{2}}-\sqrt{3}=-\frac{\sqrt{3}}{2}-4 \end{gathered}$$

Đáp án C

$$\begin{gathered} \sin (B+C)=\sin (\pi -A)=\sin A \\ \sin \frac{A+B}{2}=\sin \left(\frac{\pi }{2}-\frac{C}{2}\right)=\cos \frac{C}{2} \\ \cos (3A+B+C)=\cos (2A+\pi )=-\cos 2A \\ \sin \frac{B+C}{2}=\sin \left(\frac{\pi }{2}-\frac{A}{2}\right)=\cos \frac{A}{2} \end{gathered}$$

Đáp án B

$$\begin{gathered} \sin \frac{7\pi }{12}=\sin \left(\frac{\pi }{3}+\frac{\pi }{4}\right) \\ =\sin \frac{\pi }{3}\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\cos \frac{\pi }{3} \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}+\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4} \\ \cos {285}^0=\cos \left({360}^0-{75}^0\right)=\cos {75}^0=\cos \left({30}^0+{45}^0\right) \\ =\cos {30}^0\cos {45}^0-\sin {30}^0\sin {45}^0 \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \\ \sin \frac{\pi }{12}=\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right) \\ =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\sin \frac{\pi }{4}\cos \frac{\pi }{3} \\ =\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4} \\ \sin \frac{103\pi }{12}=\sin \left(8\pi +\frac{7\pi }{12}\right)=\sin \frac{7\pi }{12}=\frac{\sqrt{6}+\sqrt{2}}{4} \end{gathered}$$

Đáp án C

$$\begin{gathered} {\sin }^{2}x.{\tan }^{2}x+4{\sin }^{2}x-{\tan }^{2}x+3{\cos }^{2}x \\ =({\sin }^{2}x-1).{\tan }^{2}x+4{\sin }^{2}x+3{\cos }^{2}x \\ =-{\cos }^{2}x.{\tan }^{2}x+4{\sin }^{2}x+3{\cos }^{2}x \\ =-{\sin }^{2}x+4{\sin }^{2}x+3(1-{\sin }^{2}x)=3 \end{gathered}$$