Câu hỏi:

Ta có:

• $\cos(-108^{\circ}) = \cos(108^{\circ}) = -\cos(72^{\circ})$


• $\cot(72^{\circ}) = \frac{\cos(72^{\circ})}{\sin(72^{\circ})}$


• $\tan(-162^{\circ}) = -\tan(162^{\circ}) = \tan(18^{\circ})$


• $\sin(108^{\circ}) = \sin(72^{\circ})$

Do đó, $A = \frac{-\cos(72^{\circ}) \cdot \frac{\cos(72^{\circ})}{\sin(72^{\circ})}}{\tan(18^{\circ}) \cdot \sin(72^{\circ})} - \tan(18^{\circ}) = \frac{-\cos^2(72^{\circ})}{\tan(18^{\circ})\sin^2(72^{\circ})} - \tan(18^{\circ})$


$= \frac{-\cos^2(72^{\circ})}{\frac{\sin(18^{\circ})}{\cos(18^{\circ})}\sin^2(72^{\circ})} - \tan(18^{\circ}) = \frac{-\cos^2(72^{\circ}) \cos(18^{\circ})}{\sin(18^{\circ})\sin^2(72^{\circ})} - \tan(18^{\circ})$


Vì $72^{\circ} = 90^{\circ} - 18^{\circ}$ nên $\sin(72^{\circ}) = \cos(18^{\circ})$ và $\cos(72^{\circ}) = \sin(18^{\circ})$.


$A = \frac{-\sin^2(18^{\circ}) \cos(18^{\circ})}{\sin(18^{\circ})\cos^2(18^{\circ})} - \tan(18^{\circ}) = -\frac{\sin(18^{\circ})}{\cos(18^{\circ})} - \tan(18^{\circ}) = -\tan(18^{\circ}) - \tan(18^{\circ}) = -2\tan(18^{\circ})$


Tuy nhiên, các đáp án không phù hợp. Có vẻ như có lỗi trong đề bài hoặc đáp án. Nếu đề bài là $A = \frac{{\cos ( - 108^\circ ).\cot 72^\circ }}{{\tan ( - 162^\circ ).\sin 108^\circ }} + \tan 18^\circ $ thì $A = -\tan(18^{\circ}) + \tan(18^{\circ}) = 0$.

Nếu đề bài là $A = \frac{{\cos ( - 108^\circ ).\cot 72^\circ }}{{\tan ( - 162^\circ ).\sin 108^\circ }} $ thì $A = -\tan(18^{\circ})$.

Nếu biểu thức là $A = \frac{{\cos ( - 108^\circ ).\cot 72^\circ }}{{\tan ( - 162^\circ ).\sin 108^\circ }} - 1$ thì $A = -\tan(18^{\circ})-1$

Xét biểu thức $A = \frac{{\cos ( - 108^\circ ).\cot 72^\circ }}{{\tan ( - 162^\circ ).\sin 108^\circ }} - \frac{1}{2}$ thì kết quả khác.
Nếu kết quả là $\frac{1}{2}$ thì biểu thức ban đầu phải là $A = \frac{{\cos ( - 108^\circ ).\cot 72^\circ }}{{\tan ( - 162^\circ ).\sin 108^\circ }} + \frac{3}{2}$
Với biểu thức đã cho, không có đáp án nào đúng.

Ta có: $A = \frac{{{{(1 - {{\tan }^2}\alpha )}^2}}}{{4{{\tan }^2}\alpha }} - \frac{1}{{4{{\sin }^2}\alpha .{{\cos }^2}\alpha }}$

$A = \frac{{{{(1 - \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }})}^2}}}{{4\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} - \frac{1}{{4{{\sin }^2}\alpha .{{\cos }^2}\alpha }}$

$A = \frac{{{{(\frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }})}^2}}}{{4\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} - \frac{1}{{4{{\sin }^2}\alpha .{{\cos }^2}\alpha }}$

$A = \frac{{{{(\cos 2\alpha )}^2}}}{{4{{\sin }^2}\alpha {{\cos }^2}\alpha }} - \frac{1}{{4{{\sin }^2}\alpha {{\cos }^2}\alpha }}$

$A = \frac{{{{\cos }^2}2\alpha - 1}}{{4{{\sin }^2}\alpha {{\cos }^2}\alpha }}$

$A = \frac{{ - {{\sin }^2}2\alpha }}{{{{\sin }^2}2\alpha }} = - \frac{1}{4}$

Ta có:
$A = \cos^2\alpha \cdot \cot^2\alpha + 3\cos^2\alpha - \cot^2\alpha + 2\sin^2 \alpha$

$A = \cos^2\alpha \cdot \frac{\cos^2\alpha}{\sin^2\alpha} + 3\cos^2\alpha - \frac{\cos^2\alpha}{\sin^2\alpha} + 2\sin^2 \alpha$

$A = \frac{\cos^4\alpha}{\sin^2\alpha} + 3\cos^2\alpha - \frac{\cos^2\alpha}{\sin^2\alpha} + 2\sin^2 \alpha$

$A = \frac{\cos^4\alpha - \cos^2\alpha}{\sin^2\alpha} + 3\cos^2\alpha + 2\sin^2 \alpha$

$A = \frac{\cos^2\alpha(\cos^2\alpha - 1)}{\sin^2\alpha} + 3\cos^2\alpha + 2\sin^2 \alpha$

$A = \frac{\cos^2\alpha(-\sin^2\alpha)}{\sin^2\alpha} + 3\cos^2\alpha + 2\sin^2 \alpha$

$A = -\cos^2\alpha + 3\cos^2\alpha + 2\sin^2 \alpha$

$A = 2\cos^2\alpha + 2\sin^2 \alpha$

$A = 2(\cos^2\alpha + \sin^2 \alpha)$

$A = 2 \cdot 1 = 2$

Ta có:

• $\tan(x) = \cot(90^\circ - x)$


• $\tan(x) \cdot \cot(x) = 1$

Do đó:
$D = (\tan 1^\circ \cdot \cot 1^\circ) \cdot (\tan 2^\circ \cdot \cot 2^\circ) \dots (\tan 89^\circ \cdot \cot 89^\circ) = 1 \cdot 1 \dots 1 = 1$.