giải PT :
tanx - sin2x - cos2x+ 2(2cosx -1/cosx )= 0
B=\(\dfrac{sin2x}{tanx+cot2x}\)
a. cho sin a + cos a = \(\frac{-1}{3}\)tính sin a .cos a
b. chứng minh đẳng thức \(\frac{sin4x}{1+cos4x}.\frac{cos2x}{1+cos2x}=tanx\)
chứng minh
a> cot2x/1+cot2x . 1+tan2x/tan2x = tan2x+cot2x/1=tan4x
b>tan2x-cos2x/sin2x + cot2x-sin2x/cos2x = 2
a) \(1-cot^4x=\frac{2}{sin^2x}-\frac{1}{sin^4x}\)
b)\(\frac{1-2sinx.cosx}{cos^2-sin^2}\)\(=\frac{1-tanx}{1+tanx}\)\(\)
c)\(\frac{sin^2x}{sinx-cosx}+\frac{sinx+cosx}{1-tanx}=sinx+cosx\)
d)\(\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}=\frac{2.cosx}{|sin|}\)
e)\(tan^3x+tan^2x+tanx+1=\frac{sinx+cosx}{cos^3x}\)
\(\text{3sinx + 2cosx = 3(1+tanx) - 1/cosx}\)
\(\sqrt{sin^2x\left(1+cotx\right)+cos^2x\left(1+tanx\right)}\)
Rút gọn giúp tui nha~~
rút gọn bt sau (3.tanx-tan^3x)/(1-3.tan^2x)
Rút gọn biểu thức sau A = (tanx + cotx)2 - ( tanx - cotx)2
A. 3
B. 4
C. 2
D. 1
