Verify the identity of #sec^6X(secXtanX)-sec^4X(secXtanX)=sec^5Xtan^3X ?#

3 Answers
Apr 12, 2018

See below

Explanation:

#sec^6x(secxtanx)-sec^4x(secxtanx)=sec^4xsecxtanx(sec^2x-1)=sec^5xtanxtan^2x=sec^5xtan^3x#

Apr 12, 2018

#sec^6X(secXtanX)-sec^4X(secXtanX)=sec^5Xtan^3X #

Taking the #LHS#,

#sec^6X(secXtanX)-sec^4X(secXtanX)#

#=>(sec^6X-sec^4X)(secXtanX)#

#=>sec^4X(sec^2X-1)(secXtanX)#

#=>sec^4X(tan^2X)(secXtanX)# #color(white)(ww# #["as " color(red)(sec^2X-1 =tan^2X) ]#

Getting the #tanX# and #secX# together,

#=>sec^5Xtan^3X = RHS#

Hence Verified ! :)

Apr 12, 2018

Please see below.

Explanation:

We know that,

#color(red)((1)sec^2theta-1=tan^2theta#

Here,

#sec^6X(secXtanX)-sec^4X(secXtanX)=sec^5Xtan^3X #

We take,

#LHS=sec^6X(color(blue)(secXtanX))-sec^4X(color(blue)(secXtanX))#

#=color(blue)(secXtanX)(sec^6X-sec^4X)#

#=color(blue)(secXtanX)sec^4X(sec^2X-1)#

#=sec^5XtanX(color(red)(sec^2X-1))...toApply(1)#

#=sec^5XtanX(color(red)(tan^2X))#

#=sec^5Xtan^3X#

#=RHS#