Sao ko thấy tham số m nào bạn nhỉ?

\(VT=\dfrac{1}{2}sin2a-tan^2a.cota+\dfrac{1}{2}tan^2a.sin2a\)

\(=\dfrac{1}{2}sin2a\left(1+tan^2a\right)-tana\)

\(=\dfrac{1}{2}sin2a.\dfrac{1}{cos^2a}-tana\)

\(=\dfrac{sina.cosa}{cos^2a}-tana=tana-tana=0\) (đpcm)

\(sina.cosa\le\frac{1}{2}\left(sin^2a+cos^2a\right)=\frac{1}{2}\)

Dấu "=" có xảy ra khi \(sina=cosa=\pm\frac{\sqrt{2}}{2}\)

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

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

b) khai triển hằng đẳng thức là ra

a) nhân tích chéo

\(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)\(\Leftrightarrow\cos^2\alpha+\sin^2\alpha=1\)(luôn đúng)

\(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}=\frac{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha-\sin^2\alpha-\cos^2\alpha+2\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)

\(=\frac{4\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}=4\)(đpcm)

a: \(\sin^2a+\cos^2a=1\)

\(\Leftrightarrow\cos^2a=1-\sin^2a=\left(1-\sin a\right)\left(1+\sin a\right)\)

hay \(\dfrac{\cos a}{1-\sin a}=\dfrac{1+\sin a}{\cos a}\)

b: \(VT=\dfrac{\left(\sin a+\cos a+\sin a-\cos a\right)\left(\sin a+\cos a-\sin a+\cos a\right)}{\sin a\cdot\cos a}\)

\(=\dfrac{2\cdot\cos a\cdot2\sin a}{\sin a\cdot\cos a}=4\)

Chứng minh đẳng thức nhé các bạn !!! Mình quên ghi đầu bài

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)

mjk trả lời mà bạn ko đúng cái nào

x^2+x+1

=x²+2x.1/2+1/4+3/4

=(x+1/2)²+3/4>0 với mọi x (vì (x+1/2)²\(\ge\)0)

vậy x^2+x+1>0