a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).

b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).

c) \(\dfrac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha}\)\(=\dfrac{4-4cos2\alpha+cos4\alpha-1}{4+4cos2\alpha+cos4\alpha-1}\)
\(=\dfrac{4\left(1-cos2\alpha\right)-2sin^22\alpha}{4\left(1+cos2\alpha\right)-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-2sin^22\alpha}{4sin^2\alpha-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-8sin^2\alpha cos^2\alpha}{4sin^2\alpha-8sin^2\alpha cos^2\alpha}\)
\(=\dfrac{4cos^2\alpha\left(1-2sin^2\alpha\right)}{4sin^2\alpha\left(1-2cos^2\alpha\right)}=cot^2\alpha.\dfrac{cos2\alpha}{-cot2\alpha}\)
\(=-cot^2\alpha\).

Điều kiện: a>45 độ

Lời giải:

\(\frac{\tan 2a}{\tan 4a-\tan 2a}=\frac{\tan 2a}{\frac{2\tan 2a}{1-\tan ^22a}-\tan 2a}=\frac{1}{\frac{2}{1-\tan ^22a}-1}=\frac{1-\tan ^22a}{1+\tan ^22a}\)

\(=\frac{1-\frac{\sin ^22a}{\cos ^22a}}{1+\frac{\sin ^22a}{\cos ^22a}}=\frac{\cos ^22a-\sin ^22a}{\sin ^22a+\cos ^22a}=\cos ^22a-\sin ^22a=\cos 4a\)

Ta có đpcm.

Nhân cả tử và mẫu của phân số chứa tan với \(sina.cosa\)

\(A=\frac{sin^2x-cos^2x}{sin^2x+cos^2x}+cos2x=sin^2x-cos^2x+cos2x=-cos2x+cos2x=0\)

\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}=\frac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin4a.cos4a+2cos^22a-1}\)

\(B=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(C=\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a-1\right)}{2\left(cos^22a+2cos2a+1\right)}\)

\(C=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{sin^4a}{cos^4a}=tan^4a\)

\(D=\frac{sin^22a+4sin^4a-\left(2sina.cosa\right)^2}{4-4sin^2a-sin^22a}=\frac{sin^22a+4sin^4a-sin^22a}{4\left(1-sin^2a\right)-\left(2sina.cosa\right)^2}=\frac{4sin^4a}{4cos^2a-4sin^2a.cos^2a}\)

\(=\frac{sin^4a}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^2a.cos^2a}=\frac{sin^4a}{cos^4a}=tan^4a\)

a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).

b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).

c) \(\dfrac{1+tan^4\alpha}{tan^2\alpha+cot\alpha}=\left(1+tan^4\alpha\right):\left(tan^2\alpha+cot^2\alpha\right)\)
\(=\left(1+\dfrac{sin^4\alpha}{cos^4\alpha}\right):\left(\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}\right)\)
\(=\dfrac{sin^4\alpha+cos^4\alpha}{cos^4\alpha}:\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) (Đpcm).