`h)` Ta có:
`VP=1/ 2 ({cosx}/{1+sinx}+{cosx}/{1-sinx})` $(ĐK: sinx\ne ±1)$
`=1/ 2 cosx (1/{1+sinx}+1/{1-sinx})`
`=1/ 2 . cosx. {1-sinx+1+sinx}/{(1+sinx)(1-sinx)}`
`=1/ 2 cosx . 2/{1-sin^2 x}`
`=1/ 2 .cosx . 2/{cos^2 x}`
`=1/{cosx}=VT`
Vậy:
`1/{cosx}=1/ 2 ({cosx}/{1+sinx}+{cosx}/{1-sinx})` $(đpcm)$
`k)`
Ta có:
`VT=1+tanx+tan^2 x+tan^3 x` $(ĐK: cosx\ne 0)$
`= (tan^3 x+tan x)+(tan^2 x+1)`
`=tanx (tan^2 x+1)+(tan^2 x+1)`
`=(tan^ 2 x+1)(tanx+1)`
`=({sin^2 x}/{cos^2 x}+1)({sinx}/{cosx}+1)`
`={sin^2 x+cos^2 x}/{cos^2 x} . {sinx+cosx}/{cosx}`
`=1/{cos^2 x} . {sinx+cosx}/{cosx}`
`={sinx+cosx}/{cos^3 x}=VP`
Vậy:
`1+tanx+tan^2 x+tan^3 x={sinx+cosx}/{cos^3 x}` $(đpcm)$
