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\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
\(\frac{sin^2\alpha}{cos\alpha.\left(1+\frac{sin\alpha}{cos\alpha}\right)}-\frac{cos^2\alpha}{sin\alpha.\left(1+\frac{cos\alpha}{sin\alpha}\right)}=\frac{sin^2\alpha}{cos\alpha+sin\alpha}-\frac{cos^2\alpha}{sin\alpha+cos\alpha}=\frac{\left(sin\alpha+cos\alpha\right).\left(sin\alpha-cos\alpha\right)}{sin\alpha+cos\alpha}=sin\alpha-cos\alpha\)
Câu a chắc bạn ghi nhầm \(\frac{cota+1}{cota-1}\) thành \(\frac{cosa+1}{cota-1}\)
\(\frac{2}{tana-1}+\frac{cota+1}{cota-1}=\frac{2cota}{1-cota}+\frac{cota+1}{cota-1}=\frac{-2cota+cota+1}{cota-1}=\frac{1-cota}{-\left(1-cota\right)}=-1\)
\(2\left(sin^6x+cos^6x\right)-3\left(sin^4x+cos^4x\right)\)
\(=2\left(sin^2x+cos^2x\right)^3-6sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3\left(sin^2x+cos^2x\right)^2+6sin^2x.cos^2x\)
\(=-1-6sin^2x.cos^2x+6sin^2x.cos^2x=-1\)
oh , bác sĩ ơi tui sắp chết