\(\sin^4\alpha+\sin^2\alpha.\cos^2\alpha+\cos^2\alpha=\)\(\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha=\sin^2\alpha+\cos^2\alpha=1\)

\(\frac{1}{1+\sin\alpha}+\frac{1}{1-\sin\alpha}-2\tan^2\alpha=\frac{1-\sin\alpha+1+\sin\alpha}{1-\sin^2\alpha}-\frac{2\sin^2\alpha}{\cos^2\alpha}=\)

\(\frac{2}{1-\sin^2\alpha}-\frac{2\sin^2\alpha}{\cos^2\alpha}=2\left(\frac{1}{\cos^2\alpha}-\frac{\sin^2\alpha}{\cos^2\alpha}\right)=2\)

chúng không phụ thuộc vào số đo góc\(\alpha\)

a, Ta có tổng các góc bằng 180^o

=> \(\widehat{P}=55^o\)

- Áp dụng tỉ số lượng giác :

\(\cos35=\dfrac{MN}{4}\)

\(\Rightarrow MN\approx3,277cm\)

\(\sin35=\dfrac{MP}{4}\)

\(\Rightarrow MP\approx2,294cm\)

b, Ta có : \(A=\dfrac{2\cos^2a-\cos^2a-\sin^2a}{\sin a+\cos a}=\dfrac{\left(\sin a+\cos a\right)\left(\cos a-\sin a\right)}{\sin a+\cos a}\)

\(=\cos a-\sin a\)

c, \(sin30< sin35< cos40< sin60< cos25\)

a.Ta có \(\tan\alpha.\cot\alpha=1\Rightarrow\tan\alpha=\frac{1}{\cot\alpha}\)

\(\Rightarrow\frac{1}{\cot\alpha}+\cot\alpha=2\Rightarrow\cot^2\alpha-2\cot\alpha+1=0\)

\(\cot\alpha=1\Rightarrow\alpha=45^0\)

b.Ta có \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)

\(\Rightarrow7.\sin^2\alpha+5\left(1-\sin^2\alpha\right)=\frac{13}{2}\)\(\Leftrightarrow\sin^2\alpha=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}sin\alpha=\frac{\sqrt{3}}{2}\\sin\alpha=\frac{-\sqrt{3}}{2}\end{cases}}\)

\(\Rightarrow\alpha=60^0\)

\(A=\sin^2\alpha+\cos^2\alpha+2.\sin\alpha.\cos\alpha-2\sin\alpha.\cos\alpha-1\)

\(=1-1=0\) nên với mọi góc \(\alpha\) thì A không phụ thuộc vào \(\alpha\)