Đáp án + Giải thích các bước giải:

$\textbf{2.1}$

$\textbf{a}\bigg)$

$\begin{array}{l} \textbf{M} &= \cos^2 15^\circ+ \cos^2 25^\circ + \cos^2 35^\circ + \cos^2 45^\circ + \cos^2 55^\circ \\ & + \cos^2 65^\circ + \cos^2 75^\circ
\\ &= \cos^2 15^\circ + \cos^2 25^\circ + \cos^2 35^\circ + \cos^2 45^\circ + \cos^2 (90^\circ - 35^\circ)
\\ & + \cos^2 (90^\circ - 25^\circ) + \cos^2 (90^\circ - 15^\circ)
\\ &= \cos^2 15^\circ + \cos^2 25^\circ + \cos^2 35^\circ + \cos^2 45^\circ + \sin^2 35^\circ
\\ &+ \sin^2 25^\circ + \sin^2 15^\circ
\\ &= (\cos^2 15^\circ + \sin^2 15^\circ) + (\cos^2 25^\circ + \sin^2 25^\circ) + (\cos^2 35^\circ
\\ & + \sin^2 35^\circ) + \cos^2 45^\circ
\\ &= 1 + 1 + 1 +\bigg(\dfrac{\sqrt{2}}{2}\bigg)^2
\\ &= \dfrac{7}{2} \end{array}$ 

$\textbf{a}\bigg)$

$\begin{array}{l} \textbf{N} &= \sin^2 10^\circ - \sin^2 20^\circ + \sin^2 30^\circ - \sin^2 40^\circ - \sin^2 50^\circ \\ & - \sin^2 70^\circ + \sin^2 80^\circ
\\ &= \sin^2 10^\circ - \sin^2 20^\circ + \sin^2 30^\circ - \sin^2 40^\circ - \sin^2 (90^\circ - 40^\circ)
\\ & - \sin^2 (90^\circ - 20^\circ)  + \sin^2 (90^\circ - 10^\circ)
\\ &= \sin^2 10^\circ - \sin^2 20^\circ + \sin^2 30^\circ - \sin^2 40^\circ -  \cos^2 40^\circ
\\ &-  \cos^2 20^\circ + \cos^2 10^\circ
\\ &= (\sin^2 10^\circ + \cos^2 10^\circ) - (\sin^2 20^\circ + \cos^2 20^\circ) - (\sin^2 40^\circ
\\ & + \cos^2 40^\circ) + \sin^2 30^\circ
\\ &= 1 - 1 - 1 +\bigg(\dfrac{1}{2}\bigg)^2
\\ &= -\dfrac{3}{4} \end{array}$ 

$\textbf{2.2}$

Ta có:

$\tan \alpha = \dfrac{\sin\alpha}{\cos \alpha}$

$\cot \alpha = \dfrac{\cos \alpha}{\sin \alpha}$

Vì $\alpha$ là góc nhọn nên $\begin {cases} 0 < \sin \alpha < 1 \\ 0 < \cos \alpha < 1 \end {cases}$

Suy ra $\begin {cases} \dfrac{\sin \alpha}{\cos \alpha} > \sin \alpha \\ \dfrac{\cos \alpha}{\sin \alpha} > \cos \alpha \end {cases}$

$\begin {cases} \sin \alpha < \tan\alpha \\ \cos \alpha < \cot \alpha \end {cases}$

$\textbf{2.3}$

Ta có: $\cos^2 \alpha + \sin^2 \alpha = 1$

Suy ra $\sin^2 \alpha + 0,16 = 1$

$\sin^2 \alpha = 0,84$

$\sin \alpha = \dfrac{\sqrt{21}}{5}$

$\begin{array}{l} & \tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{\frac{\sqrt{21}}{5}}{0,4} = \dfrac{\sqrt{21}}{2}
\\ & \cot \alpha = \dfrac{\cos \alpha}{\sin \alpha} = \dfrac{1}{\cot \alpha} = \dfrac{2\sqrt{21}}{21}\end{array}$

$\textbf{2.4}$

Ta có: $\cos \alpha - \sin \alpha = \dfrac{1}{5}$

Suy ra$\sin \alpha = \cos \alpha - \dfrac{1}{5}$

Do đó $\cos^ 2\alpha + \sin^2 \alpha = \cos^2 \alpha + \bigg(\cos \alpha - \dfrac{1}{5}\bigg)^2 = 1$

$\cos^2 \alpha + \cos^2 \alpha - \dfrac{2}{5} \cos \alpha + \dfrac{1}{25} = 1$

$2 \cos^2 \alpha - \dfrac{2}{5} \cos \alpha - \dfrac{24}{25} = 0$

$50 \cos^2 \alpha - 10 \cos \alpha - 24 = 0$

$(5 \cos \alpha + 3)(10 \cos \alpha - 8) = 0$

$5 \cos \alpha + 3 = 0$ hoặc $10 \cos \alpha - 8 = 0$

$\cos \alpha = -\dfrac{3}{5}(ktm)$ hoặc $\cos \alpha = \dfrac{4}{5}(tm)$

Lúc này $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \dfrac{16}{25} = \dfrac{9}{25}$

$\sin \alpha = \dfrac{3}{5}$

Suy ra $\cot \alpha = \dfrac{\cos \alpha}{\sin \alpha} = \dfrac{\frac{4}{5}}{\frac{3}{5}} = \dfrac{4}{3}$

$\textbf{2.5}$

Xét $\triangle ABC$ vuông tại $C$, ta có:

$\cos A = \dfrac{AC}{AB} = \dfrac{5}{13}$

$\Rightarrow AC = \dfrac{5}{13}AB$

$\Rightarrow AB^2 = AC^2 + BC^2 = \dfrac{25}{169}AB^2 + BC^2$

$\Rightarrow BC^2 = \dfrac{144}{169}AB^2$

$\Rightarrow BC = \dfrac{12}{13}AB$

$\Rightarrow \tan B = \dfrac{AC}{BC} = \dfrac{\frac{5}{13}AB}{\frac{12}{13}AB} = \dfrac{5}{12}$

$\textbf{2.6}$

$\textbf{a}\bigg)$

$\begin{array}{l} \textbf{VT} &= \dfrac{\cos \alpha}{1 - \sin \alpha}
\\ &= \dfrac{\cos \alpha (1 + \sin \alpha)}{(1 + \sin \alpha)(1 - \sin \alpha)}
\\ &= \dfrac{\cos \alpha (1 + \sin \alpha)}{1 - \sin^2 \alpha}
\\ &= \dfrac{\cos \alpha (1 +\sin \alpha)}{\cos^2 \alpha}
\\ &= \dfrac{1 + \sin \alpha}{\cos \alpha}  = \textbf{VP} \end{array}$

$\textbf{b}\bigg)$

$\begin{array}{l} \textbf{VT} &= \dfrac{(\sin \alpha + \cos \alpha)^2 - (\sin \alpha-  \cos \alpha)^2}{\sin \alpha \cos \alpha} 
\\ &= \dfrac{(\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha) - (\sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha)}{\sin \alpha \cos \alpha}
\\ &= \dfrac{1 + 2\sin \alpha \cos \alpha - (1 - 2\sin \alpha \cos \alpha)}{\sin \alpha + \cos \alpha}
\\ &= \dfrac{1 + 2 \sin \alpha \cos \alpha - 1 + 2\sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha}
\\ &= \dfrac{4 \sin \alpha \cos \alpha}{\sin \alpha \cos \alpha}
\\ &= 4 = \textbf{VP}\end{array}$