chứng minh rằng:
a)\(1-\frac{sin^2a}{1+cot\text{a}}-\frac{c\text{os}^2a}{1+t\text{ana}}=sin\text{a}.c\text{os}a\)
b)\(\frac{sin^4a+c\text{os}^4-1}{sin^6a+c\text{os}^6a-1}=\frac{2}{3}\)
c)\(\frac{1+c\text{os}a}{1-c\text{os}a}-\frac{1-c\text{os}a}{1+c\text{os}a}=\frac{4cot\text{a}}{sin\text{a}}\)
d)\(\frac{c\text{os}a.cot\text{a}-sin\text{a}.t\text{ana}}{\frac{1}{sin\text{a}}-\frac{1}{c\text{os}a}}=1+sin\text{a}.c\text{os}a\)
e)\(\frac{c\text{os}a+sin\text{a}-1}{c\text{os}a-sin\text{a}+1}=\frac{sin\text{a}}{1+c\text{os}a}\)
f)\(\frac{sin\text{a}}{1+c\text{os}a}+\frac{1+c\text{os}a}{sin\text{a}}=\frac{2}{sin\text{a}}\)
a: \(VT=1-\dfrac{sin^2a}{1+\dfrac{cosa}{sina}}-\dfrac{cos^2a}{1+\dfrac{sina}{cosa}}\)
\(=1-sin^2a:\dfrac{sina+cosa}{sina}-cos^2a:\dfrac{cosa+sina}{cosa}\)
\(=1-\dfrac{sin^3a+cos^3a}{sina+cosa}\)
\(=1-sin^2a-cos^2a+sinacosa=sinacosa\)
b: \(=\dfrac{\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a-1}{\left(sin^2a+cos^2a\right)^3-3\cdot sin^2\cdot cos^2a-1}\)
\(=\dfrac{-2\cdot\left(sina\cdot cosa\right)^2}{-3\cdot\left(sina\cdot cosa\right)^2}=\dfrac{2}{3}\)
c: \(=\dfrac{1+2cosa+cos^2a-1+2cosa-cos^2a}{1-cos^2a}\)
\(=\dfrac{4cosa}{sin^2a}=\dfrac{4cota}{sina}\)
f: \(=\dfrac{sin^2a+cos^2a+1+2cosa}{sina\left(cosa+1\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\)