10 câu Trắc nghiệm Giá trị lượng giác của một góc bất kì từ 0o đến 150o có đáp án (Vận dụng)

Đáp án D

$$\begin{gathered} 7={\tan }^2\alpha +{\cot }^2\alpha \\ ={\left(\tan \alpha +\cot \alpha \right)}^2-2\tan \alpha \cot \alpha =m^2-2.1 \\ \Rightarrow m^2=9\iff m=\pm 3 \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} \cos \alpha +\sin \alpha =\frac{1}{3}\Rightarrow {\left(\cos \alpha +\sin \alpha \right)}^2=\frac{1}{9} \\ \iff 1+2\cos \alpha \sin \alpha =\frac{1}{9} \\ \iff \cos \alpha \sin \alpha =-\frac{4}{9} \end{gathered}$$

Ta có:

$$\begin{gathered} P=\sqrt{{\tan }^2\alpha +{\cot }^2\alpha } \\ =\sqrt{{\left(\tan \alpha +\cot \alpha \right)}^2-2\tan \alpha \cot \alpha } \\ =\sqrt{{\left(\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\right)}^2-2} \\ =\sqrt{{\left(\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }\right)}^2-2} \\ =\sqrt{{\left(\frac{1}{\sin \alpha \cos \alpha }\right)}^2-2} \\ =\sqrt{{\left(-\frac{9}{4}\right)}^2-2}=\frac{7}{4} \end{gathered}$$

Đáp án D

Ta có:

$$\begin{gathered} E=2{\cos }^2\alpha +5\sin \alpha \cos \alpha +1 \\ ={\sin }^2\alpha \left(2.\frac{{\cos }^2\alpha }{{\sin }^2\alpha }+5\frac{\cos \alpha }{\sin \alpha }+\frac{1}{{\sin }^2\alpha }\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(2.{\cot }^2\alpha +5\cot \alpha +\frac{1}{\frac{1}{{\cot }^2\alpha +1}}\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(2.{\cot }^2\alpha +5\cot \alpha +{\cot }^2\alpha +1\right) \\ =\left(\frac{1}{{\cot }^2\alpha +1}\right)\left(3{\cot }^2\alpha +5\cot \alpha +1\right) \\ =\frac{1}{5^2+1}\left({3.5}^2+5.5+1\right)=\frac{101}{26} \end{gathered}$$

Đáp án B

Ta có:

$$\begin{gathered} E=\frac{\cot \alpha +3\tan \alpha }{2\cot \alpha +\tan \alpha }=\frac{1+3{\tan }^2\alpha }{2+{\tan }^2\alpha } \\ =\frac{3({\tan }^2\alpha +1)-2}{1+(1+{\tan }^2\alpha )}=\frac{\frac{3}{{\cos }^2\alpha }-2}{\frac{1}{{\cos }^2\alpha }+1} \\ =\frac{3-2{\cos }^2\alpha }{1+{\cos }^2\alpha }=\frac{19}{13} \end{gathered}$$

Đáp án A

Ta có:

$$\begin{gathered} 3\cos \alpha -\sin \alpha =1\iff 3\cos \alpha =\sin \alpha +1 \\ \Rightarrow 9{\cos }^2\alpha ={\left(\sin \alpha +1\right)}^2 \\ \iff 9{\cos }^2\alpha ={\sin }^2\alpha +2\sin \alpha +1 \\ \iff 9\left(1-{\sin }^2\alpha \right)={\sin }^2\alpha +2\sin \alpha +1 \\ \iff 10{\sin }^2\alpha +2\sin \alpha -8=0 \\ \iff \left[\begin{array}{c}\sin \alpha =-1\\ \sin \alpha =\frac{4}{5}\end{array}\right. \end{gathered}$$

• $\sin \alpha =-1$ không thỏa mãn vì $0^0<\alpha <{90}^0$
• $\sin \alpha =\frac{4}{5}\Rightarrow \cos \alpha =\frac{3}{5}\Rightarrow \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{4}{3}$