Rút gọn các biểu thức :
a/ \(\frac{2cos^2\alpha-1}{sin\alpha+cos\alpha}\)
b/ \(\frac{sin25+cos70}{sin20+cos65}\)
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\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)
\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
b) \(\frac{\sin25+\cos70}{\sin20+\cos65}\)
xét tam giác vuông có : sin a= cos b => cos 70 = sin (90 -70) <=> cos 70 = sin 20
cos 65 =sin 25
<=> \(\frac{\sin25+\cos70}{\sin20+\cos65}\)
=\(\frac{\sin25+\sin20}{\sin20+\sin25}=1\)
\(\frac{2\cos^2\cdot a-1}{\sin a+\cos a}=\frac{2\cos^2a-\left(\sin^2+\cos^2\right)}{\sin a+\cos a}\)
vì \(\sin^2a+\cos^2a=1\)
=\(\frac{\cos^2a-\sin^2a}{\sin a+\cos a}=\frac{\left(\cos a-\sin a\right)\left(\cos a+\sin a\right)}{\sin a+\cos a}\)
=\(\cos a-\sin a\)
\(a,=\frac{2cos^2\alpha-cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =cos\alpha-sin\alpha\)
\(b,sin25=cos65;cos70=sin20;Khiđó:B=1\)