1)
a) cot2 α+ 1 = \(\dfrac{1}{sin^2a}\)
b)1 + tan2 α = \(\dfrac{1}{cos^2a}\)
c) sin4 α+ cos2α = 2.sin2α . cos2 α
d) \(\dfrac{1-4.sin^2a.cos^2a}{\left(sina+cosa\right)^2}=1-2.sina.cosa\)
e) \(\dfrac{2.sina.cosa-1}{cos^2a-sin^2a}=\dfrac{tana-1}{tana+1}\)
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Đọc tiếp
1)
a) cot2 α+ 1 = \(\dfrac{1}{sin^2a}\)
b)1 + tan2 α = \(\dfrac{1}{cos^2a}\)
c) sin4 α+ cos2α = 2.sin2α . cos2 α
d) \(\dfrac{1-4.sin^2a.cos^2a}{\left(sina+cosa\right)^2}=1-2.sina.cosa\)
e) \(\dfrac{2.sina.cosa-1}{cos^2a-sin^2a}=\dfrac{tana-1}{tana+1}\)
Lời giải:
a) \(\cot ^2a+1=\left(\frac{\cos a}{\sin a}\right)^2+1=\frac{\cos ^2a+\sin ^2a}{\sin ^2a}=\frac{1}{\sin ^2a}\)
b)
\(\tan ^2a+1=\left(\frac{\sin a}{\cos a}\right)^2+1=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}=\frac{1}{\cos ^2a}\)
c) Đề bài sai.
\(\sin ^4a+\cos ^2a=\sin ^2a.\sin ^2a+\cos ^2a\)
\(=\sin ^2a(1-\cos ^2a)+\cos ^2a\)
\(\sin ^2a+\cos ^2a-\sin ^2a\cos ^2a=1-\sin ^2a\cos ^2a\)
d)
\(\frac{1-4\sin ^2a\cos ^2a}{(\sin a+\cos a)^2}=\frac{1-(2\sin a\cos a)^2}{\sin ^2a+2\sin a\cos a+\cos ^2a}=\frac{(1-2\sin a\cos a)(1+2\sin a\cos a)}{1+2\sin a\cos a}\)
\(=1-2\sin a\cos a\)
e) ĐK tồn tại tan là $\cos x\neq 0$
Vì \(\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\tan a\cos a\)
Ta có:
\(\frac{2\sin a\cos a-1}{\cos ^2a-\sin ^2a}=\frac{1-2\sin a\cos a}{\sin ^2a-\cos ^2a}=\frac{\cos ^2a+\sin ^2a-2\sin a\cos a}{(\sin a-\cos a)(\sin a+\cos a)}\)
\(=\frac{(\sin a-\cos a)^2}{(\sin a-\cos a)(\sin a+\cos a)}=\frac{\sin a-\cos a}{\sin a+\cos a}\)
\(=\frac{\tan a\cos a-\cos a}{\tan a\cos a+\cos a}=\frac{\cos a(\tan a-1)}{\cos a(\tan a+1)}\)\(=\frac{\tan a-1}{\tan a+1}\) (đpcm)