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\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x+cot^2x-2=9\Rightarrow tan^2x+cot^2x=11\)
\(tan^2x+cot^2x+2=13\Rightarrow\left(tanx+cotx\right)^2=13\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=\left(tan^2x+cot^2x\right)\left(tanx-cotx\right)\left(tanx+cotx\right)\)
\(=11.3.\left(\pm\sqrt{13}\right)=\pm33\sqrt{13}\)
\(C=\frac{tan^210}{tan^2\left(90-80\right)}+\frac{tan^220}{tan^2\left(90-70\right)}+...+\frac{tan^240}{tan^2\left(90-50\right)}+tan^245\)
\(=\frac{tan^210}{tan^210}+\frac{tan^220}{tan^220}+\frac{tan^230}{tan^230}+\frac{tan^240}{tan^240}+1\)
\(=1+1+1+1+1=5\)
+) ta có : \(A=tan5.tan10...tan85\)
\(=\left(tan5.tan85\right).\left(tan10.tan80\right)...\left(tan40.tan50\right).tan45\)
\(=\left(tan5.tan\left(90-5\right)\right).\left(tan10.tan\left(90-10\right)\right)...\left(tan40.tan\left(90-40\right)\right).tan45\)
\(=\left(tan5.cot5\right).\left(tan10.cot10\right)...\left(tan40.cot40\right).tan45\)\(=tan45=1\)
+) ta có : \(B=cot3.cot6...cot87\)
\(=\left(cot3.cot87\right).\left(cot6.cot84\right)...\left(cot42.cot48\right).cot45\)
\(=\left(cot3.cot\left(90-3\right)\right).\left(cot6.cot\left(90-6\right)\right)...\left(cot42.cot\left(90-42\right)\right).cot45\)\(=\left(cot3.tan3\right).\left(cot6.tan6\right)...\left(cot42.tan42\right).cot45\)
\(=cot45=1\)
1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)
=1
2: \(sin^4x-cos^4x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)
\(=1-2\cdot cos^2x\)
\(x^2+x+\sqrt{x^2+x+1}=1\)
ĐK:....
\(pt\Leftrightarrow x^2+x+\sqrt{x^2+x+1}-1=0\)
\(\Leftrightarrow x\left(x+1\right)+\dfrac{x^2+x+1-1}{\sqrt{x^2+x+1}+1}=0\)
\(\Leftrightarrow x\left(x+1\right)+\dfrac{x\left(x+1\right)}{\sqrt{x^2+x+1}+1}=0\)
\(\Leftrightarrow x\left(x+1\right)\left(1+\dfrac{1}{\sqrt{x^2+x+1}+1}\right)=0\)
Dễ thấy: \(1+\dfrac{1}{\sqrt{x^2+x+1}+1}>0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(=cot^2x\left(cos^2x-1\right)+cos^2x+4\left(sin^2x+cos^2x\right)\)
\(=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+4\)
\(=-cos^2x+cos^2x+4=4\)
Khỏi tick
\(a,\tan10.\tan11......\)
\(=\left(\tan10.tan80\right)\left(tan11.tan79\right)....\left(tan44.tan46\right).tan45\)
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 : \(\tan^3x+\cot^3x=\left(\tan x+\cot x\right)\left(\tan^2x-\tan x\cot x+\cot^2x\right)=2\)