Chứng minh rằng tanx + cotx -sin2x = 2(sin^4x + cos^4x) / sin2x 23/08/2021 Bởi Hadley Chứng minh rằng tanx + cotx -sin2x = 2(sin^4x + cos^4x) / sin2x
Ta có: `VT=tanx+cotx-sin2x` `={sinx}/{cosx}+{cosx}/{sinx}-sin2x` `={sin^2x+cos^2x}/{sinxcosx}-sin2x` $=\dfrac{1}{\dfrac{1}{2}sin2x}-sin2x$ `=2/{sin2x}-sin2x` $\ (1)$ $\\$ `VP={2(sin^4x+cos^4x)}/{sin2x}` `={2[(sin^2x+cos^2x)^2-2sin^2xcos^2x]}/{sin2x}` $=\dfrac{2.(1-\dfrac{1}{2}sin^2 2x)}{sin2x}$ `=2/{sin2x}-{sin^2 2x}/{sin2x}` `=2/{sin2x}-sin2x` $\ (2)$ $\\$ Từ `(1);(2)` suy ra: `tanx+cotx-sin2x={2(sin^4x+cos^4x)}/{sin2x}` Bình luận
Ta có:
`VT=tanx+cotx-sin2x`
`={sinx}/{cosx}+{cosx}/{sinx}-sin2x`
`={sin^2x+cos^2x}/{sinxcosx}-sin2x`
$=\dfrac{1}{\dfrac{1}{2}sin2x}-sin2x$
`=2/{sin2x}-sin2x` $\ (1)$
$\\$
`VP={2(sin^4x+cos^4x)}/{sin2x}`
`={2[(sin^2x+cos^2x)^2-2sin^2xcos^2x]}/{sin2x}`
$=\dfrac{2.(1-\dfrac{1}{2}sin^2 2x)}{sin2x}$
`=2/{sin2x}-{sin^2 2x}/{sin2x}`
`=2/{sin2x}-sin2x` $\ (2)$
$\\$
Từ `(1);(2)` suy ra:
`tanx+cotx-sin2x={2(sin^4x+cos^4x)}/{sin2x}`