\(H=cot15^o.cot35^o.cot55^o.cot75^o\)

\(=\left(cot15^o.cot75^o\right).\left(cot35^o.cot55^o\right)\)

\(=\left(cot15^o.tan15^o\right).\left(cot35^o.tan35^o\right)\)

\(=1\)

\(I=tan10^o.tan20^o.tan30^o....tan80^o\)

\(=\left(tan10^o.cot10^o\right).\left(tan20^o.cot20^o\right).\left(tan30^o.cot30^o\right).\left(tan40^o.cot40^o\right)\)

\(=1\)

Ta có \(2\sin x\cos x=\left(\sin x+\cos x\right)^2-\left(\sin^2x+\cos^2x\right)\) 

\(=\left(\dfrac{3}{4}\right)^2-1=-\dfrac{7}{16}\)  

Từ đó \(A=\left|\sin x-\cos x\right|\)

\(\Rightarrow A^2=\left(\sin x-\cos x\right)^2\)

\(A^2=\sin^2x+\cos^2x-2\sin x\cos x\)

\(A^2=1+\dfrac{7}{16}=\dfrac{23}{16}\)

\(\Rightarrow A=\dfrac{\sqrt{23}}{4}\) (do \(A\ge0\))

Có \(\cos x+\sin x=\dfrac{3}{4}\)

\(\Leftrightarrow\left(\cos x+\sin x\right)^2=\dfrac{9}{16}\)

\(\Leftrightarrow2.\sin x.\cos x+1=\dfrac{9}{16}\)

\(\Leftrightarrow\sin x.\cos x=-\dfrac{7}{32}\)

Lại có \(\left(\cos x+\sin x\right)^2=\left(\cos x-\sin x\right)^2+4.\sin x.\cos x=\dfrac{9}{16}\)

\(\Leftrightarrow\left(\cos x-\sin x\right)^2=\dfrac{23}{16}\)

\(\Leftrightarrow\left|\sin x-\cos x\right|=\dfrac{\sqrt{23}}{4}\)

Chọn A.

Áp dụng công thức biến đổi tổng thành tích và công thức nhân đôi; ta có

Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

\(\dfrac{2sina+cosa}{2sin^3a-cos^3a}=\dfrac{\dfrac{2sina}{cos^3a}+\dfrac{cosa}{cos^3a}}{\dfrac{2sin^3a}{cos^3a}-\dfrac{cos^3a}{cos^3a}}=\dfrac{2tana.\dfrac{1}{cos^2a}+\dfrac{1}{cos^2a}}{2tan^3a-1}\)

\(=\dfrac{2tana\left(1+tan^2a\right)+1+tan^2a}{2tan^3a-1}=...\) (thay số và bấm máy)

Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)

\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)

Chọn D.

Ta có : sin2a = 2.sina. cosa và sin²a = 1 - cos²a.

Do đó;