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\(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)\(\Leftrightarrow\cos^2\alpha+\sin^2\alpha=1\)(luôn đúng)
\(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}=\frac{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha-\sin^2\alpha-\cos^2\alpha+2\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
\(=\frac{4\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}=4\)(đpcm)
Chia cả tử và mẫu cho \(cosa\)
\(D=\dfrac{\dfrac{cosa}{cosa}+\dfrac{sina}{cosa}}{\dfrac{cosa}{cosa}-\dfrac{sina}{cosa}}=\dfrac{1+tana}{1-tana}=\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}=3\)
\(=\frac{\left(\sin a+\cos a-\sin a+\cos a\right)\left(\sin a+\cos a+\sin a-\cos a\right)}{\sin a.\cos a}=\frac{2.\cos a.2.\sin a}{\sin a.\cos a}=4\)
a) Có: `1+tan^2a=1/(cos^2a)`
`<=> 1+(3/5)^2=1/(cos^2a)`
`=> cosa=\sqrt10/4`
`=> sina = \sqrt(1-cos^2a) = \sqrt6/4`
b) Có: `sin^2a + cos^2a=1`
`<=> sin^2a + (1/4)^2=1`
`=> sina=\sqrt15/4`
`=> tana = (sina)/(cosa) = \sqrt15`
Má ơi,tính sai:
a)\(\left[{}\begin{matrix}cos\alpha=\dfrac{5\sqrt{34}}{34}\\cos\alpha=\dfrac{-5\sqrt{34}}{34}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}sin\alpha=cos\alpha.tan\alpha=\dfrac{3\sqrt{34}}{34}\\sin\alpha=cos\alpha.tan\alpha=\dfrac{-3\sqrt{34}}{34}\end{matrix}\right.\)
b)\(\left[{}\begin{matrix}sin\alpha=\dfrac{\sqrt{15}}{4}\\sin\alpha=\dfrac{-\sqrt{15}}{4}\end{matrix}\right.\)\(\left[{}\begin{matrix}tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\sqrt{15}\\tatn\alpha=-\sqrt{15}\end{matrix}\right.\)
a: \(\sin^2a+\cos^2a=1\)
\(\Leftrightarrow\cos^2a=1-\sin^2a=\left(1-\sin a\right)\left(1+\sin a\right)\)
hay \(\dfrac{\cos a}{1-\sin a}=\dfrac{1+\sin a}{\cos a}\)
b: \(VT=\dfrac{\left(\sin a+\cos a+\sin a-\cos a\right)\left(\sin a+\cos a-\sin a+\cos a\right)}{\sin a\cdot\cos a}\)
\(=\dfrac{2\cdot\cos a\cdot2\sin a}{\sin a\cdot\cos a}=4\)