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\(a,\tan10.\tan11......\)
\(=\left(\tan10.tan80\right)\left(tan11.tan79\right)....\left(tan44.tan46\right).tan45\)
Mà 10 và 80, 11 và 79, ... là các góc phụ nhau .
\(=tan10.cot10....tan45=1\)
b, Ta có : \(\tan x+\cot x=2\)
\(\Rightarrow\tan^2x+2\tan x\cot x+\cot^2x=4\)
\(\Rightarrow\tan^2x+\cot^2x=4-2=2\)
Ta có : \(\tan^3x+\cot^3x=\left(\tan x+\cot x\right)\left(\tan^2x-\tan x\cot x+\cot^2x\right)=2\)
Có \(\sin^2x+\cos^2x=1\Rightarrow\sin^2x-\cos^2x=1-2\cos^2x\)
\(\Rightarrow VT=\frac{\sin^2x-\cos^2x}{\sin^2x.\cos^2x}=\frac{\sin^4x-\cos^4x}{\sin^2x.\cos^2x}=\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=\tan^2x-\cot^2x=VP\)
\(=cot^2x\left(cos^2x-1\right)+cos^2x+4\left(sin^2x+cos^2x\right)\)
\(=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+4\)
\(=-cos^2x+cos^2x+4=4\)
Khỏi tick
\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)
\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)
\(=1-\cot^4x\)=VT
a) A xác định khi \(\left\{{}\begin{matrix}x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt{x}\ne3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)
b)Với \(x>0;x\ne9\), ta có:
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để A đạt giá trị nguyên thì \(\frac{4}{\sqrt{x}-3}\) đạt giá trị nguyên
Hay\(4⋮\left(\sqrt{x}-3\right)\)
Suy ra \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
TH1: \(\sqrt{x}-3=\pm1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=1\\\sqrt{x}-3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=4\end{matrix}\right.\)
TH2: \(\sqrt{x}-3=\pm2\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=2\\\sqrt{x}-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\\x=1\end{matrix}\right.\)
TH3: \(\sqrt{x}-3=\pm4\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=4\\\sqrt{x}-3=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=7\\\sqrt{x}=-1\left(Loại\right)\end{matrix}\right.\Rightarrow x=49\)
Vậy \(x\in\left\{1;4;16;25;49\right\}\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x+cot^2x-2=9\Rightarrow tan^2x+cot^2x=11\)
\(tan^2x+cot^2x+2=13\Rightarrow\left(tanx+cotx\right)^2=13\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=\left(tan^2x+cot^2x\right)\left(tanx-cotx\right)\left(tanx+cotx\right)\)
\(=11.3.\left(\pm\sqrt{13}\right)=\pm33\sqrt{13}\)