Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có \(\sin^2x+\cos^2x=1\Rightarrow\sin^2x-\cos^2x=1-2\cos^2x\)
\(\Rightarrow VT=\frac{\sin^2x-\cos^2x}{\sin^2x.\cos^2x}=\frac{\sin^4x-\cos^4x}{\sin^2x.\cos^2x}=\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=\tan^2x-\cot^2x=VP\)
\(=cot^2x\left(cos^2x-1\right)+cos^2x+4\left(sin^2x+cos^2x\right)\)
\(=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+4\)
\(=-cos^2x+cos^2x+4=4\)
Khỏi tick
\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)
\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)
\(=1-\cot^4x\)=VT
1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)
=1
2: \(sin^4x-cos^4x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)
\(=1-2\cdot cos^2x\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x+cot^2x-2=9\Rightarrow tan^2x+cot^2x=11\)
\(tan^2x+cot^2x+2=13\Rightarrow\left(tanx+cotx\right)^2=13\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=\left(tan^2x+cot^2x\right)\left(tanx-cotx\right)\left(tanx+cotx\right)\)
\(=11.3.\left(\pm\sqrt{13}\right)=\pm33\sqrt{13}\)