Question

[Nam Truong Van]

Điền vào chỗ trống {......}để đơn giản các biểu thức sau:

a)\(\dfrac{ }{ }\) 1 + tan\(^2\) a =1 +\((\dfrac{...}{...})\)² =\(\dfrac{....+....}{cos^2a}=\dfrac{........}{cos^2a}\)

b) 1 + cot² a= + \((\dfrac{...}{...})^2\) = \(\dfrac{....+....}{sin^2a}=\dfrac{....}{sin^2a}\)

c) tan² a (2 sin²a + 3 cos² a - 2)

=tan² a[cos² a +2 (........ +.........)-2 ]

=\(\dfrac{sin^2a}{cos^2a}\)\(\times\)........=........

Answer 1

[lâm mỹ ngọc]

a) 1 + tan a =1 + =\(\dfrac{sina+cosa}{cos^2a}\)

b) 1 + cot2 a= 1 +(\(\dfrac{cosa}{sina}\))² = \(\dfrac{cosa+sina}{sin^2a}\)=\(\dfrac{1}{sin^2a}\)

c) tan2 a (2 sin2a + 3 cos2 a - 2)

=tan2 a[cos2 a +2 (\(sina^2+cos^2a\))-2 ]

=\(\dfrac{sin^2a}{cos^2a}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

b: \(1+cot^2a=1+\left(\dfrac{cosa}{sina}\right)^2=\dfrac{1}{sin^2a}\)

c: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)

\(=tan^2a\left[cos^2a\right]\)

\(=\dfrac{sin^2a}{cos^2a}\cdot cos^2a=sin^2a\)