How do you prove $sec^6x(secxtanx) - sec^4x(secxtanx) = sec^5xtan^3x$ ?

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Answer 1 · 2018-03-11T05:19:17

$LHS=sec^6x(secxtanx) - sec^4x(secxtanx)$

$=sec^5xtanx(sec^2x - 1)$

$=sec^5xtanx*tan^2x$

$= sec^5xtan^3x=RHS$

Answer 2 · 2018-03-11T08:28:30

As proved.

To prove $sec^6x(secs tanx) - sec^4x(sec x tan x)$ = sec^5x tan^3x#

$L H S = sec ^6x (sec x tan x) - sec^4x(sec x tan x)$

$=> sec x tan x) ( sec ^6x - sec ^4x)$ taking common term $color(brown)(sec x tan x$ out.

$=> (sec x tan x) * (sec^4x) (sec^2x - 1)$ taking $color(brown)(sec^4x$ out.

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$=> (sec x tan x)(sec ^4x) (tan^2x)$ as $sec^2x - 1 = tan^2x$

$=> (sec x * sec^4x) * (tan x * tan^2 x)$ Rearranging like terms together.

$=> sec^5x * tan ^3x = R H S$

Q E D

How do you prove sec^6x(secxtanx) - sec^4x(secxtanx) = sec^5xtan^3xsec^6x(secxtanx) - sec^4x(secxtanx) = sec^5xtan^3x?