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\(VT=\left[\frac{\left(1+sinx\right)-\left(1-sinx\right)}{\sqrt{1-sin^2x}}\right]^2=\left(\frac{2sinx}{cosx}\right)^2=4tan^2x=VP\left(đpcm\right)\)
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\(cosx=\sqrt{1-\dfrac{7}{16}}=\dfrac{3}{4}\)
\(tanx=\dfrac{\sqrt{7}}{4}:\dfrac{3}{4}=\dfrac{\sqrt{7}}{3}\)
\(cotx=1:\dfrac{\sqrt{7}}{3}=\dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{7}\)
\(M=\left(\dfrac{3}{7}\sqrt{7}+\dfrac{1}{3}\sqrt{7}\right):\left(\dfrac{3}{7}\sqrt{7}-\dfrac{1}{3}\sqrt{7}\right)\)
\(=\dfrac{16}{21}:\dfrac{2}{21}=8\)
a/ Tớ làm bên dưới rồi
b/ \(\frac{1}{sin^2x}=\frac{sin^2x+cos^2x}{sin^2x}=\frac{\frac{sin^2x}{sin^2x}+\frac{cos^2x}{sin^2x}}{\frac{sin^2x}{sin^2x}}=1+cot^2x\)(đpcm)
c/ \(\frac{1}{tanx+1}+\frac{1}{cotx+1}=\frac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}=\frac{tanx+cotx+2}{tanx.cotx+tanx+cotx+1}\)
\(=\frac{tanx+cotx+2}{tanx+cotx+2}=1\left(đpcm\right)\)
d/ \(\frac{tan^2x-cos^2x}{sin^2x}+\frac{cot^2x-sin^2x}{cos^2x}=\frac{tan^2x}{sin^2x}-\frac{cos^2x}{sin^2x}+\left(\frac{cot^2x}{cos^2x}-\frac{sin^2x}{cos^2x}\right)\)
\(=\frac{\frac{sin^2x}{cos^2x}}{sin^2x}-\frac{cos^2x}{sin^2x}+\frac{\frac{cos^2x}{sin^2x}}{cos^2x}-\frac{sin^2x}{cos^2x}\)
\(=\frac{1}{cos^2x}-cot^2x+\frac{1}{sin^2x}-tan^2x\)
\(=1+tan^2x-cot^2x+\left(1+cot^2x\right)-tan^2x\)
\(=1+tan^2x-cot^2x+1+cot^2x-tan^2x=2\left(đpcm\right)\)
a) ta có : \(sin^2x+cos^2x=1\Leftrightarrow\dfrac{9}{25}+cos^2x=1\Leftrightarrow cos^2x=\dfrac{16}{25}\)
\(\Rightarrow cosx=\pm\dfrac{4}{5}\)
ta có : \(tanx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{3}{5}}{\pm\dfrac{4}{5}}=\pm\dfrac{3}{4}\) \(\Rightarrow cot=\dfrac{1}{tan}=\dfrac{1}{\pm\dfrac{3}{4}}=\pm\dfrac{4}{3}\)
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b) ta có : \(tanx=\sqrt{3}\Leftrightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\sqrt{3}}\)
ta có : \(\dfrac{sin^2x+cos^2x}{cos^2x}=1+tan^2x\Leftrightarrow\dfrac{1}{cos^2x}=1+tan^2x\)
\(\Leftrightarrow\dfrac{1}{cos^2x}=1+\left(\sqrt{3}\right)^2=4\Rightarrow cos^2x=\dfrac{1}{4}\) \(\Leftrightarrow cos^2x=\pm\dfrac{1}{2}\)
ta có : \(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\Rightarrow sinx=\pm\dfrac{\sqrt{3}}{2}\)
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câu c bn làm tương tự câu a ; còn câu d bn làm tương tự câu b nha :)