Tính \(\tan32^O.\cot32^O-\sin45^O.\cos45^O\)
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\(D=\cos45^0\cdot\cos^223^0+\sin45^0\cdot\cos^267^0\)
\(=\dfrac{\sqrt{2}}{2}\left(\cos^223^0+\cos^267^0\right)\)
\(=\dfrac{\sqrt{2}}{2}\)
\(B=\sin^247^o\times\cos45^o+\sin45^o\times\cos^247^o\)
\(B=\sin^247^o\times\cos45^o+\cos45^o\times\cos^247^o\)
\(B=\cos45^o\left(\sin^247^o+\cos^247^o\right)\)
\(B=\cos45^o.1=\cos45^o\)
\(\sin^225^o+\sin^265^o-\tan35^o+\cot55^o-\frac{\cot32^o}{tan58^o}\)
\(=\cos^265^o+\sin^265^o-\cot55^{^{ }o}+\cot55^o-\frac{\tan58^o}{\tan58^o}\)
\(=1-0-1\)
\(=0\)
nhớ k cho mik nha ^^
Ta có \(\cot\alpha=\tan\beta\) ; \(\cos^2\alpha+\sin^2\alpha=1\)
Khi đó \(-\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}+1=\frac{-\cot58^{\text{o}}-\tan27^{\text{o}}+\cot63^{\text{o}}+\tan32^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}\)
\(=\frac{\left(\tan32^{\text{o}}-\cot58^{\text{o}}\right)+\left(\cot63^{\text{o}}-\tan27^{\text{o}}\right)}{\cot63^{\text{o}}+\tan32^{\text{o}}}=0\)
=> \(\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}=1\)
=> \(\cos^255^{\text{o}}-\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}=\cos^255^{\text{o}}-1=-\sin^255\)
\(A=sin^225+cos^2\left(90-65\right)-tan35+tan\left(90-55\right)-\frac{cot32}{cot\left(90-58\right)}\)
\(=sin^225+cos^225-tan35+tan35-\frac{cot32}{cot32}\)
\(=1-0-1=0\)