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\(tan10^0.tan80^0.tan20^0.tan70^0.tan30.tan60.tan40.tan50\)
\(=tan10.tan\left(90-10\right).tan20.tan\left(90-20\right).tan30.tan\left(90-30\right).tan40.tan\left(90-40\right)\)
\(=tan10.cot10.tan20.cot20.tan30.cot30.tan40.cot40\)
\(=1.1.1.1=1\)
\(tan19.tan33+tan33.tan38+tan38.tan19\)
\(=tan33\left(tan19+tan38\right)+tan38.tan19\)
\(=\frac{sin33}{cos33}\left(\frac{sin19}{cos19}+\frac{sin38}{cos38}\right)+\frac{sin38.sin19}{cos38.cos19}\)
\(=\frac{sin33}{cos33}\left(\frac{sin19.cos38+sin38.cos19}{cos19.cos38}\right)+\frac{sin38.sin19}{cos38.cos19}\)
\(=\frac{sin33}{cos33}.\frac{sin57}{cos19.cos38}+\frac{sin38.sin19}{cos38.cos19}=\frac{sin33}{cos33}.\frac{cos33}{cos19.cos38}+\frac{\frac{1}{2}\left(cos19-cos57\right)}{cos38.cos19}\)
\(=\frac{2sin33-cos19-cos57}{2cos38.cos19}=\frac{2sin33-cos19-sin33}{2cos38.cos19}=\frac{sin33-cos19}{2cos38.cos19}\)
\(=\frac{cos\left(90-33\right)-cos19}{cos57-cos19}=\frac{cos57-cos19}{cos57-cos19}=1\)
Bài giải đã giải thích rồi mà......Với 0<t<1 =>\(\left\{\begin{matrix}t^3>0\\1-t>0\end{matrix}\right.\) tích hai số dương => phải dương
Sửa đề: \(2\cdot sin\left(180-a\right)\cdot cota-cos\left(180-a\right)\cdot tana+cot\left(180-a\right)\)
\(=2\cdot sina\cdot cota+cosa\cdot tana+\dfrac{cos\left(180-a\right)}{sin\left(180-a\right)}\)
\(=2\cdot sina\cdot\dfrac{cosa}{sina}+cosa\cdot\dfrac{sina}{cosa}+\dfrac{-cosa}{sina}\)
\(=2cosa+sina-tana\)
S= (cos100+cos1700) + (cos300+cos1500) + (cos500+cos1300)+(cos700+1100)+cos900
=0
\(B=sin20-sin80+sin40\)
\(B=-2cos50.sin30+sin40\)
\(B=-cos50+sin40\)
\(B=-cos\left(90-40\right)+sin50\)
\(B=-sin40+sin40=0\)
\(C=sin160-sin100+sin\left(180-40\right)\)
\(C=2cos130.sin30+sin40\)
\(C=cos130+sin40\)
\(C=cos\left(90+40\right)+sin40\)
\(C=-sin40+sin40=0\)