Question

[Thành Danh Đỗ]

Giúp em bài 595, 596 và 597 với ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

595:

a: \(\dfrac{cota\cdot cotb+1}{cota\cdot cotb-1}=\left(\dfrac{cosa}{sina}\cdot\dfrac{cosb}{sinb}+1\right):\left(\dfrac{cosa}{sina}\cdot\dfrac{cosb}{sinb}-1\right)\)

\(=\dfrac{cosa\cdot cosb+sina\cdot sinb}{sina\cdot sinb}:\dfrac{cosa\cdot cosb-sina\cdot sinb}{sina\cdot sinb}\)

\(=\dfrac{cos\left(a-b\right)}{cos\left(a+b\right)}\)

3: \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{cos^2a\cdot cos^2b}=\dfrac{\left(sina\cdot cosb+sinb\cdot cosa\right)\left(sina\cdot cosb-sinb\cdot cosa\right)}{cos^2a\cdot cos^2b}\)

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

\(=\dfrac{sin^2a\cdot cos^2b-cos^2a\left(1-cos^2b\right)}{cos^2a\cdot\left(cos^2b\right)}\)

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

tan^2a-tan^2b

=(sin^2a/cos^2a)-(sin^2b/cos^2b)

=(\(=\dfrac{\left(sina\cdot cosb\right)^2-\left(sinb\cdot cosa\right)^2}{cos^2a\cdot cos^2b}=\dfrac{\left(sina\cdot cosb-sinb\cdot cosa\right)\left(sina\cdot cosb+sinb\cdot cosa\right)}{cos^2a\cdot cos^2b}\)

=sin(a+b)sin(a-b)/cos^2a*cos^2b

4:

1-tan^2a*tan^2b

\(=1-\dfrac{sin^2a\cdot sin^2b}{cos^2a\cdot cos^2b}=\dfrac{cos^2a\cdot cos^2b-sin^2a\cdot sin^2b}{cos^2a\cdot cos^2b}\)

\(=\dfrac{\left(cosa\cdot cosb-sina\cdot sinb\right)\left(cosa\cdot cosb+sina\cdot sinb\right)}{cos^2a\cdot cos^2b}\)

\(=\dfrac{cos\left(a+b\right)\cdot cos\left(a-b\right)}{cos^2a\cdot cos^2b}\)