Lời giải

Biến đổi $\cot \frac{89\pi }{6}=\cot \left(-\frac{\pi }{6}+15\pi \right)$

$=\cot \left(-\frac{\pi }{6}\right)=-\cot \frac{\pi }{6}=-\sqrt{3}$

Lời giải

Biến đổi 

$$\begin{gathered} \tan {180}^{\circ }=\tan \left(0^{\circ }+{180}^{\circ }\right) \\ =\tan 0^{\circ }=0 \end{gathered}$$

Lời giải

Vì $\frac{\pi }{2}<a<\pi$$\Rightarrow \sin a>0$,$\cos a<0$

Lời giải

Vì $2\pi <a<\frac{5\pi }{2}$

$\Rightarrow \tan a>0,\cot a>0$

Lời giải

$$\begin{gathered} A=\frac{3\sin \alpha +\cos \alpha }{\sin \alpha -\cos \alpha } \\ =\frac{3\tan \alpha +1}{\tan \alpha -1}=7 \end{gathered}$$

Lời giải

B đúng vì:

$$\begin{gathered} {\sin }^2\alpha +{\text{cos}}^2\alpha \\ ={\left(\frac{1}{2}\right)}^2+{\left(-\frac{\sqrt{3}}{2}\right)}^2=1 \end{gathered}$$

Lời giải

Ta có: ${\sin }^2\alpha =1-{\cos }^2\alpha$

$$\begin{gathered} =1-{\left(\frac{4}{5}\right)}^2=\frac{9}{25} \\ \Rightarrow \sin \alpha =\pm \frac{3}{5} \end{gathered}$$

Do $0<\alpha <\frac{\pi }{2}$ nên $\sin \alpha >0$.

Suy ra, $\sin \alpha =\frac{3}{5}$

Lời giải

$A=\left(1–{\sin }^{2}x\right).{\cot }^{2}x+\left(1–{\cot }^{2}x\right)$

$$\begin{gathered} =c{\mathrm{ot}}^{2}x-{\cos }^{2}x+1-{\cot }^{2}x \\ ={\sin }^{2}x \end{gathered}$$

Lời giải

Theo công thức.

Lời giải

$$\begin{gathered} D={\cos }^{2}x.co{t}^{2}x \\ +3{\cos }^{2}x–co{t}^{2}x+2{\sin }^{2}x \\ ={\cos }^{2}x+2+{\cot }^{2}x\left({\cos }^{2}x-1\right) \\ ={\cos }^{2}x+2-{\cot }^{2}x.{\sin }^{2}x \\ ={\cos }^{2}x+2-{\cos }^{2}x=2 \end{gathered}$$

Lời giải

$$\begin{gathered} A=\frac{2}{{\sin }^{2}x-\sin x.\cos x-{\cos }^{2}x} \\ =\frac{\frac{2}{{\sin }^{2}x}}{1-\cot x-{\cot }^{2}x} \\ =\frac{2\left(1+{\cot }^{2}x\right)}{1-\cot x-{\cot }^{2}x} \\ =\frac{2\left(1+\frac{1}{4}\right)}{1-\frac{1}{2}-\frac{1}{4}}=10. \end{gathered}$$

Lời giải

$$\begin{gathered} A=\frac{\sin \left(-{328}^0\right).\sin {958}^0}{\cot {572}^0} \\ -\frac{\cos \left(-{508}^0\right).\cos \left(-{1022}^0\right)}{\tan \left(-{212}^0\right)} \\ \iff A=-\frac{\sin {32}^0.\sin {58}^0}{\cot {32}^0} \\ -\frac{\cos {32}^0.\cos {58}^0}{\tan {32}^0} \\ A=-\frac{\sin {32}^0.\cos {32}^0}{\cot {32}^0}-\frac{\cos {32}^0.sin{32}^0}{\tan {32}^0} \\ =-{\sin }^2{32}^0-{\cos }^2{32}^0=-1. \end{gathered}$$

Lời giải

$$\begin{gathered} A=\frac{\cos {30}^0+\sin {60}^0}{\sin {30}^0-\cos {30}^0} \\ =\frac{2\sqrt{3}}{1-\sqrt{3}}=-3-\sqrt{3} \end{gathered}$$

Lời giải

$A=\sin \alpha +\cos \alpha +\sin \alpha -\cos \alpha$

Lời giải

$$\begin{gathered} \cot 1458°=\cot \left(4.360°+18°\right) \\ =\cot 18°=\sqrt{5+2\sqrt{5}} \end{gathered}$$

Lời giải

Ta có  ${\sin }^{4}x-{\cos }^{4}x=\frac{98}{81}-A$

$$\begin{gathered} \iff \cos 2x=A-\frac{98}{81} \\ 5\left({\sin }^{4}x+{\cos }^{4}x\right)=\frac{98}{81}+A \\ \iff 1-\frac{1}{2}{\sin }^{2}2x=\frac{1}{5}\left(\frac{98}{81}+A\right) \\ \iff \frac{1}{2}+\frac{1}{2}{\cos }^{2}2x=\frac{1}{5}\left(\frac{98}{81}+A\right) \\ \iff 1+{\left(A-\frac{98}{81}\right)}^2=\frac{2}{5}\left(A+\frac{98}{81}\right) \\ =\frac{2}{5}\left(A-\frac{98}{81}\right)+\frac{392}{405} \end{gathered}$$

Đặt  $A-\frac{98}{81}=t$

$$\begin{gathered} \Rightarrow t^2-\frac{2}{5}t+\frac{13}{405}=0 \\ \iff \left[\begin{array}{l}t=\frac{13}{45}\\ t=\frac{1}{9}\end{array}\right. \end{gathered}$$

+)  $t=\frac{13}{45}\Rightarrow A=\frac{607}{405}$

+) $t=\frac{1}{9}\Rightarrow A=\frac{107}{81}.$

Lời giải

$$\begin{gathered} \sin x+\cos x=\frac{1}{2} \\ \Rightarrow {\left(\sin x+\cos x\right)}^2=\frac{1}{4} \\ \iff 2\sin x.\cos x=-\frac{3}{4} \\ \Rightarrow \sin x.\cos x=-\frac{3}{8} \end{gathered}$$

Khi đó $\sin x,\cos x$ là nghiệm của phương trình:

$$\begin{gathered} X^2-\frac{1}{2}X-\frac{3}{8}=0 \\ \Rightarrow \left[\begin{array}{l}\sin x=\frac{1+\sqrt{7}}{4}\\ \sin x=\frac{1-\sqrt{7}}{4}\end{array}\right. \end{gathered}$$

Ta có $\sin x+\cos x=\frac{1}{2}$

$\Rightarrow 2\left(\sin x+\cos x\right)=1$

+) Với  $\sin x=\frac{1+\sqrt{7}}{4}$

$\Rightarrow 3\sin x+2\cos x=\frac{5+\sqrt{7}}{4}$

+) Với $\sin x=\frac{1-\sqrt{7}}{4}$

$\Rightarrow 3\sin x+2\cos x=\frac{5-\sqrt{7}}{4}$

Lời giải

$$\begin{gathered} A=a{\cos }^{2}x+2b\sin x.cosx+c{\sin }^{2}x \\ \iff \frac{A}{{\cos }^{2}x}=a+2b\tan x+c{\tan }^{2}x \\ \iff A\left(1+{\tan }^{2}x\right) \\ =a+2b\tan x+c{\tan }^{2}x \\ \iff A\left(1+{\left(\frac{2b}{a-c}\right)}^2\right) \\ =a+2b\frac{2b}{a-c}+c{\left(\frac{2b}{a-c}\right)}^2 \\ \iff A\frac{{\left(a-c\right)}^2+{\left(2b\right)}^2}{{\left(a-c\right)}^2} \\ =\frac{a{\left(a-c\right)}^2+4b^2\left(a-c\right)+c4b^2}{{\left(a-c\right)}^2} \\ \iff A\frac{{\left(a-c\right)}^2+{\left(2b\right)}^2}{{\left(a-c\right)}^2} \\ =\frac{a{\left(a-c\right)}^2+4b^{2}a}{{\left(a-c\right)}^2} \\ =\frac{a.\left({\left(a-c\right)}^2+4b^2\right)}{{\left(a-c\right)}^2} \\ \iff A=a \end{gathered}$$

Lời giải

$$\begin{gathered} A=\frac{{\tan }^{2}a-{\sin }^{2}a}{{\cot }^{2}a-{\cos }^{2}a} \\ \iff A=\frac{{\sin }^{2}a\left(\frac{1}{{\cos }^{2}a}-1\right)}{{\cos }^2\left(\frac{1}{{\sin }^{2}a}-1\right)} \\ =\frac{{\tan }^{2}a.{\tan }^{2}a}{{\cot }^{2}a}={\tan }^{6}a \end{gathered}$$