Question #6ad0a

1 Answer
Jan 19, 2017

Given #cscx-sinx=a#

#=>1/sinx-sinx=a#

#=>(1-sin^2x)/sinx=a#

#=>cos^2x/sinx=a#

Again #secx-cosx=b#

#=>1/cosx-cosx=b#

#=>(1-cos^2x)/cosx=b#

#=>sin^2x/cosx=b#

Now #a^2b^2(a^2+b^2+3)#

#=cos^4x/sin^2x xxsin^4x/cos^2x(cos^4x/sin^2x+sin^4x/cos^2x+3)#

#=sin^2x xxcos^2x((cos^6x+sin^6x)/(sin^2xcos^2x)+3)#

#=cos^6x+sin^6x+3sin^2xcos^2x#

#=(sin^2x)^3+(cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x)#

#=(sin^2x+cos^2x)^3=1^3=1#

Proved