Please solve q 14 ?

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1 Answer
May 2, 2018

Given that tan^2x=1-a^2

rarrsecx=sqrt(1+tan^2x)=sqrt(1+1-a^2)=sqrt(2-a^2)=(2-a^2)^(1/2)

LHS=secx+tan^3x*cscx

=secx+cancel(sinx)/cosx*(sin^2x/cos^2x)*1/cancel(sinx)

=secx+sec^3x*sin^2x=secx(1+sec^2x*sin^2x)

=secx(1+sin^2x/cos^2x)=secx*[(sin^2x+cos^2x)/cos^2x]

=sec^3x=((2-a^2)^((1/2)))^3=(2-a^2)^(3/2)=RHS