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\(\tan x=\frac{\sin x}{\cos x}=\frac{3}{5}\Rightarrow\sin x=\frac{3}{5}\cos x\)
\(\Rightarrow N=\frac{\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\sin x.\cos x}{\left(\sin x-\cos x\right)\left(\sin x+\cos x\right)}\)
\(=\frac{\frac{3}{5}.\cos^2x}{\left(\frac{3}{5}\cos x-\cos x\right)\left(\frac{3}{5}\cos x+\cos x\right)}=\frac{\frac{3}{5}\cos^2x}{\frac{-16}{25}.\cos^2x}=\frac{-15}{16}\)
a) sin = đối / huyền => sinx < 1 => sinx - 1 < 0
b) cos = kề / huyền => cosx < 1 => 1 - cosx > 0
c) sinx - cosx = sinx - sin(90-x)
Nếu x > 90-x hay x > 45 thì sinx - sin(90-x) > 0 hay sinx - cosx > 0
Nếu x < 90-x hay x < 45 thì sinx - sin(90-x) < 0 hay sinx - cosx < 0
d) Tương tự câu c)
\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
\(=sinx.cosx+\dfrac{sin^2x}{\dfrac{cosx+sinx}{sinx}}+\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}\)
\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)
\(=sinx.cosx+\dfrac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{cosx+sinx}\)
\(=sinx.cosx+sin^2x+cos^2x-sinx.cosx\)
\(=sin^2x+cos^2x=1\)
a) \(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{\sqrt{\left(1+\cos x\right)^2}-\sqrt{\left(1-\cos x\right)^2}}{\sqrt{\left(1-\cos x\right)\left(1+\cos x\right)}}\)
\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)
b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)
\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)
c) (ko bt có sai đề ko, làm mãi ko ra)
d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)
\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)
\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)
\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)
\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)
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 :)
