Toán Cho $t = tanx-cotx$, tính $sin^{2}x.$$cos^{2}x$ theo $t$ 07/09/2021 By Brielle Cho $t = tanx-cotx$, tính $sin^{2}x.$$cos^{2}x$ theo $t$
Đáp án: `sin^2x.cos^2x=\frac{1}{t^2+4}` Giải: Ta có: `t=tanx-cotx` ⇒ `t^2=tan^2x-2tanx.cotx+cot^2x` ⇔ `t^2=tan^2+cot^2-2.1` ⇔ `t^2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-2` ⇔ `t^2+2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}` ⇔ `t^2+2=\frac{sin^4x+cos^4x}{sin^2x.cos^2x}` ⇔ `t^2+2=\frac{(sin^2x+cos^2x)^2-2sin^2x.cos^2x}{sin^2x.cos^2x}` ⇔ `t^2+2=\frac{1-2sin^2x.cos^2x}{sin^2x.cos^2x} \ (1)` Đặt `sin^2x.cos^2x=a` `(1) ⇔ t^2+2=\frac{1-2a}{a}` ⇔ `t^2a+2a=1-2a` ⇔ `(t^2+4)a=1` ⇔ `a=\frac{1}{t^2+4}` Vậy `sin^2x.cos^2x=\frac{1}{t^2+4}` Trả lời
Đáp án: `sin^2x.cos^2x=\frac{1}{t^2+4}`
Giải:
Ta có:
`t=tanx-cotx`
⇒ `t^2=tan^2x-2tanx.cotx+cot^2x`
⇔ `t^2=tan^2+cot^2-2.1`
⇔ `t^2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-2`
⇔ `t^2+2=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}`
⇔ `t^2+2=\frac{sin^4x+cos^4x}{sin^2x.cos^2x}`
⇔ `t^2+2=\frac{(sin^2x+cos^2x)^2-2sin^2x.cos^2x}{sin^2x.cos^2x}`
⇔ `t^2+2=\frac{1-2sin^2x.cos^2x}{sin^2x.cos^2x} \ (1)`
Đặt `sin^2x.cos^2x=a`
`(1) ⇔ t^2+2=\frac{1-2a}{a}`
⇔ `t^2a+2a=1-2a`
⇔ `(t^2+4)a=1`
⇔ `a=\frac{1}{t^2+4}`
Vậy `sin^2x.cos^2x=\frac{1}{t^2+4}`