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\(B=tan^210.tan^280.tan^220.tan^270.tan^230.tan^260.tan^240.tan^250\)
\(=\left(tan10.cot10\right)^2.\left(tan20.cot20\right)^2...\left(tan40.cot40\right)^2\)
\(=1.1.1....1=1\)
+) 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 : \(B=cot3.cot6...cot87\)
\(=\left(cot3.cot87\right).\left(cot6.cot84\right)...\left(cot42.cot48\right).cot45\)
\(=\left(cot3.cot\left(90-3\right)\right).\left(cot6.cot\left(90-6\right)\right)...\left(cot42.cot\left(90-42\right)\right).cot45\)\(=\left(cot3.tan3\right).\left(cot6.tan6\right)...\left(cot42.tan42\right).cot45\)
\(=cot45=1\)
c: \(\cot50^0>\cos50^0>\cos70^0\)
a: \(\tan40^0>\cos40^0>\cos60^0\)
b: \(\cot70^0=\tan20^0>\sin20^0>\sin10^0\)
\(A=2sin30-2cos60+tan45=2\cdot\frac{1}{2}-2\cdot\frac{1}{2}+1=1\)
\(B=\left(cot46.cot44\right)\cdot cot45=\left(cot46\cdot tan46\right)\cdot cot45=1\cdot1=1\)
\(A=2.\frac{1}{2}-2.\frac{1}{2}+1=1\)
\(B=\tan46^o.\cot46^o.\cot45^o=1.1=1\)
Bài 1 :
\(C=cos^2a\left(cos^2a+sin^2a\right)+sin^2a=cos^2a+sin^2a=1\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x+cot^2x-2=9\Rightarrow tan^2x+cot^2x=11\)
\(tan^2x+cot^2x+2=13\Rightarrow\left(tanx+cotx\right)^2=13\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=\left(tan^2x+cot^2x\right)\left(tanx-cotx\right)\left(tanx+cotx\right)\)
\(=11.3.\left(\pm\sqrt{13}\right)=\pm33\sqrt{13}\)
\(C=\frac{tan^210}{tan^2\left(90-80\right)}+\frac{tan^220}{tan^2\left(90-70\right)}+...+\frac{tan^240}{tan^2\left(90-50\right)}+tan^245\)
\(=\frac{tan^210}{tan^210}+\frac{tan^220}{tan^220}+\frac{tan^230}{tan^230}+\frac{tan^240}{tan^240}+1\)
\(=1+1+1+1+1=5\)