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a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha\)= 3 \(\sin\alpha\)
ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)
#mã mã#
a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)
\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)
\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
a) ta có : \(A=tan1.tan2.tan3...tan89\)
\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)
\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)
\(=\left(tan1.cot1\right).\left(tan2.cot2\right).\left(tan3.cot3\right)...\left(tan44.cot44\right).tan45\) \(=tan45=1\)b) ta có \(B=\dfrac{sin\alpha+2cos\alpha}{3sin\alpha-4cos\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}\)
\(=\dfrac{tan\alpha+2}{3tan\alpha-4}=\dfrac{\dfrac{1}{2}+2}{\dfrac{3}{2}-4}=-1\)
ta có \(D=\dfrac{2sin^2\alpha-3cos^2\alpha}{4cos^2\alpha-5sin^2\alpha}=\dfrac{\dfrac{2sin^2\alpha}{cos^2\alpha}-\dfrac{3cos^2\alpha}{cos^2\alpha}}{\dfrac{4cos^2\alpha}{cos^2\alpha}-\dfrac{5sin^2\alpha}{cos^2\alpha}}\)
\(=\dfrac{2tan^2\alpha-3}{4-5tan^2\alpha}=\dfrac{2\left(\dfrac{1}{2}\right)^2-3}{4-5\left(\dfrac{1}{2}\right)^2}=\dfrac{-10}{11}\)
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)