5a) \(\dfrac{cos\alpha}{1-sin\alpha}=\dfrac{1+sin\alpha}{cos\alpha}\Rightarrow cos^2\alpha=1-sin^2\alpha\Rightarrow cos^2\alpha+sin^2\alpha=1\)
Giả sử tam giác ABC vuông tại A.
Ta có: \(cos^2B+sin^2B=\dfrac{AB^2}{BC^2}+\dfrac{AC^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)
\(\Rightarrow\) đpcm
b) \(\dfrac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha.cos\alpha}\)
\(=\dfrac{sin^2\alpha+cos^2\alpha+2sin\alpha.cos\alpha-sin^2\alpha-cos^2\alpha+2sin\alpha.cos\alpha}{sin\alpha.cos\alpha}\)
\(=\dfrac{4sin\alpha.cos\alpha}{sin\alpha.cos\alpha}=4\)
6.a) Ta có \(cos\alpha=sin\left(90-\alpha\right)\)
\(\Rightarrow\left\{{}\begin{matrix}cos28=sin62\\cos41=sin49\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}cos^228=sin^262\\cos^241=sin^249\end{matrix}\right.\)
\(\Rightarrow A=sin^262+cos^262+sin^249+cos^249=1+1=2\) (vì \(sin^2\alpha+cos^2\alpha=1\))
b) Ta có: \(sin^2\alpha+cos^2\alpha=1\Rightarrow sin^2\alpha+cos^2\alpha-2sin\alpha.cos\alpha=0\)
\(\Rightarrow\left(sin\alpha-cos\alpha\right)^2=0\Rightarrow sin\alpha=cos\alpha\)
Giả sử tam giác ABC vuông tại A có \(sinB=cosB\)
\(\Rightarrow\dfrac{AC}{BC}=\dfrac{AB}{BC}\Rightarrow AB=AC\Rightarrow\Delta ABC\) vuông cân tại A
\(\Rightarrow\angle B=45\)
c) Ta có: \(cot\alpha=\dfrac{1}{tan\alpha}\Rightarrow tan\alpha+\dfrac{1}{tan\alpha}=2\)
\(\Rightarrow\dfrac{tan^2\alpha+1}{tan\alpha}=2\Rightarrow tan^2\alpha+1=2tan\alpha\Rightarrow tan^2\alpha-2tan\alpha+1=0\)
\(\Rightarrow\left(tan\alpha-1\right)^2=0\Rightarrow tan\alpha=1\)
Giả sử tam giác ABC vuông tại A có \(tanB=1\)
\(\Rightarrow\dfrac{AC}{AB}=1\Rightarrow AB=AC\Rightarrow\Delta ABC\) vuông cân tại A
\(\Rightarrow\angle B=45\)
Bài 5:
a) Ta có: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)
\(\Leftrightarrow1-\sin^2\alpha=\cos^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)
b) Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}=4\)
\(\Leftrightarrow4\sin\alpha\cdot\cos\alpha=4\sin\alpha\cdot\cos\alpha\)
