Please solve q 14 ?

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Answer 1 · 2018-05-02T18:22:07

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#