\(\dfrac{sina}{sin^3a+\cos^3a}\)
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\(tana=\sqrt{3}\)
nên \(\dfrac{sina}{cosa}=\sqrt{3}\)
=>\(sina=\sqrt{3}\cdot cosa\)
=>a=60 độ
\(A=\dfrac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina\cdot cosa\right)}{sina-cosa}\)
\(=1+sina\cdot cosa=1+\dfrac{1}{2}sin2a\)
\(=1+\dfrac{1}{2}\cdot sin120=\dfrac{4+\sqrt{3}}{4}\)
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ta có : \(sin^3a+cos^3a=\left(sina+cosa\right)^3-3sina.cosa\left(sina+cosa\right)\)
\(=2^3-3sina.cosa\left(2\right)=8-6sina.cosa\)
\(=11-3sin^2a-6sina.cosa-3cos^2a=11-3\left(sin+cos\right)^2=11-3.2^2=11-12=-1\)
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\(A=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}+sina+cosa\)
\(=1+sina.cosa+sina+cosa\)
\(=\left(sina+1\right)\left(cosa+1\right)\)
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Lời giải:
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(1+\frac{\cos a}{\sin a})\sin ^3a+(1+\frac{\sin a}{\cos a})\cos ^3a\)
\(=(\sin a+\cos a)\sin ^2a+(\cos a+\sin a)\cos ^2a\)
\(=(\sin a+\cos a)(\sin ^2a+\cos ^2a)=(\sin a+\cos a).1=\sin a+\cos a\)
Đề yêu cầu gì bạn nhỉ?