Đề kiểm tra Giá trị lượng giác của một góc từ 0 độ đến 180 độ lớp 10 (có lời giải) - Đề 1

Chọn C

\(E = \frac{{\cot \alpha  - 3\tan \alpha }}{{2\cot \alpha  - \tan \alpha }} = \frac{{1 - 3{{\tan }^2}\alpha }}{{2 - {{\tan }^2}\alpha }} = \frac{{4 - 3\left( {{{\tan }^2}\alpha  + 1} \right)}}{{3 - \left( {1 + {{\tan }^2}\alpha } \right)}} = \frac{{4 - \frac{3}{{{{\cos }^2}\alpha }}}}{{3 - \frac{1}{{{{\cos }^2}\alpha }}}} = \frac{{4{{\cos }^2}\alpha  - 3}}{{3{{\cos }^2}\alpha  - 1}} =  - \frac{{11}}{3}\).

Chọn C

\(\begin{array}{l}{\left( {\cot a + \tan a} \right)^2} = {\cot ^2}a + 2\cot a.\tan a + {\tan ^2}a\\ = \left( {{{\cot }^2}a + 1} \right) + \left( {{{\tan }^2}a + 1} \right) = \frac{1}{{{{\sin }^2}a}} + \frac{1}{{{{\cos }^2}a}}\end{array}\).

Chọn C

\(A = \frac{{{{\cot }^2}x - {{\cos }^2}x}}{{{{\cot }^2}x}} + \frac{{\sin x.\cos x}}{{\cot x}} = \frac{{\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - {{\cos }^2}x}}{{\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}} + \frac{{\sin x.\cos x}}{{\frac{{\cos x}}{{\sin x}}}}\)\( = \frac{{{{\cos }^2}x\left( {1 - {{\sin }^2}x} \right)}}{{{{\cos }^2}x}} + {\sin ^2}x = 1 - {\sin ^2}x + {\sin ^2}x = 1\).

Vì $\alpha \in \left({90}^{°};{180}^{°}\right)$ nên \(\sin \alpha  > 0\).

Ta có: \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\sin ^2}\alpha  = 1 - {\cos ^2}\alpha  = 1 - \frac{4}{9} = \frac{5}{9}\) mà \(\sin \alpha  > 0\),

nên \(\sin \alpha  = \frac{{\sqrt 5 }}{3};\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{ - \frac{2}{3}}}{{\frac{{\sqrt 5 }}{3}}} =  - \frac{2}{{\sqrt 5 }},\tan \alpha  =  - \frac{{\sqrt 5 }}{2}\).

Vì \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {\left( {\frac{2}{3}} \right)^2} = \frac{5}{9}\).

Mà $0^{°}<\alpha <{90}^{°}$ nên \(\cos \alpha  > 0\). Do đó \(\cos \alpha  = \frac{{\sqrt 5 }}{3}\) nên\(A = \frac{{\sin \alpha  + \sqrt 5 \cos \alpha }}{{2\sin \alpha  + \cos \alpha }} = \frac{7}{{4 + \sqrt 5 }}\).

a) \(D = a.1 + b.0 + c.0 = a\).

b) \(E = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 2 \cdot {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - 5.1 =  - \frac{{11}}{4}\).

c) \(F = {\cos ^2}{24^0} + {\sin ^2}{24^0} + {\cos ^2}{1^0} + {\sin ^2}{1^0} = 1 + 1 = 2\).

d) $G={\cos }^2{45}^{°}+{\sin }^2{45}^{°}+{\sin }^2{45}^{°}-2\left({\cos }^2{50}^{°}+{\cos }^2{40}^{°}\right)-5\tan {55}^{°}\cot {55}^{°}$

$={\cos }^2{45}^{°}+{\sin }^2{45}^{°}+{\sin }^2{45}^{°}-2\left({\cos }^2{50}^{°}+{\sin }^2{50}^{°}\right)-5\tan {55}^{°}\cot {55}^{°}=\frac{-11}{2}$

Vì \(\cot \alpha  =  - 3\)nên \(\sin \alpha  \ne 0.\) Chia cả tử và mẫu của \(P\) cho \({\sin ^3}\alpha \), ta có:

\(P = \frac{{1 + {{\cot }^3}\alpha }}{{\frac{1}{{{{\sin }^2}\alpha }}\left( {1 - \cot \alpha } \right)}} = \frac{{1 + {{\cot }^3}\alpha }}{{\left( {1 + {{\cot }^2}\alpha } \right)\left( {1 - \cot \alpha } \right)}} = \frac{{1 + {{\left( { - 3} \right)}^3}}}{{\left[ {1 + {{\left( { - 3} \right)}^2}} \right]\left[ {1 - \left( { - 3} \right)} \right]}} =  - \frac{{13}}{{20}}\)

Do \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{3} \Rightarrow \sin \alpha  \ne 0\).

Chia hai vế biểu thức \(A\) cho \(\sin \alpha \), ta có: \(A = \frac{{3 + 4\frac{{\cos \alpha }}{{\sin \alpha }}}}{{2 - 5\frac{{\cos \alpha }}{{\sin \alpha }}}} = \frac{{3 + 4\cot \alpha }}{{2 - 5\cot \alpha }} = \frac{{3 + 4 \cdot \frac{1}{3}}}{{2 - 5 \cdot \frac{1}{3}}} = 13\)

Ta có: \(\sin a + \cos a = \sqrt 2 \) \( \Rightarrow 2 = {(\sin a + \cos a)^2} = {\sin ^2}a + 2\sin a\cos a + {\cos ^2}a = 1 + 2\sin a\cos a\) \( \Rightarrow \sin a \cdot \cos a = \frac{1}{2}\).

Khi đó: \({\sin ^4}a + {\cos ^4}a = \left( {{{\sin }^2}a + {{\cos }^2}a} \right) - 2{\sin ^2}a{\cos ^2}a = 1 - 2{\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}\).

Vì \(\tan \alpha  = 1 \Rightarrow \cos \alpha  \ne 0\). Chia cả tử và mẫu cho \({\cos ^2}\alpha \) ta được:

\(B = \frac{{\left( {{{\sin }^2}\alpha  + 1} \right)\frac{1}{{{{\cos }^2}\alpha }}}}{{\left( {2{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\frac{1}{{{{\cos }^2}\alpha }}}} = \frac{{{{\tan }^2}\alpha  + \frac{1}{{{{\cos }^2}\alpha }}}}{{2 - {{\tan }^2}\alpha }} = \frac{{{{\tan }^2}\alpha  + {{\tan }^2}\alpha  + 1}}{{2 - {{\tan }^2}\alpha }} = 3\)