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#

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