Question

Prove that sec^6θ=tan^6θ+3tan^2θsec^2θ+1

Answer 1

Use the identities,

$1+\tan^2 \theta =\sec^2 \theta$

$(a^2+b^2)^3=a^6+3a^2b^2(a^2+b^2)+b^6$

$LHS=\sec^6 \theta =(\sec^2 \theta)^3\\ LHS=\sec^6 \theta =(1+\tan^2 \theta)^3\\ LHS=\sec^6 \theta =\tan^6 \theta+3\tan^2 \theta(1+\tan^2 \theta)+1\\ LHS=\sec^6 \theta =\tan^6 \theta+3\tan^2 \theta \sec^2 \theta+1=RHS$