Question

[Sách Giáo Khoa]

Chứng minh các đẳng thức :

a) \(\dfrac{\tan\alpha-\tan\beta}{\cot\beta-\cot\alpha}=\tan\alpha\tan\beta\)

b) \(\tan100^0+\dfrac{\sin530^0}{1+\sin640^0}=\dfrac{1}{\sin10^0}\)

c) \(2\left(\sin^6\alpha+\cos^6\alpha\right)+1=3\left(\sin^4\alpha+\cos^4\alpha\right)\)

Answer 1

[Bùi Thị Vân]

a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).

Answer 2

[Bùi Thị Vân]

b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .