Question 946977
{{{tan^2(x)+6=sec^2(x)+5}}}.........since {{{tan^2(x)=sin^2(x)/cos^2(x)}}}, we have

{{{sin^2(x)/cos^2(x)+6=sec^2(x)+5}}} ...........since {{{sec^2(x)=1/cos^2(x)}}}, we have


{{{sin^2(x)/cos^2(x)+6=1/cos^2(x)+5}}}


{{{sin^2(x)/cos^2(x)-1/cos^2(x)=5-6}}}


{{{(sin^2(x)-1)/cos^2(x)=-1}}}


{{{sin^2(x)-1=-1*cos^2(x)}}} ..........since {{{sin^2(x)=1-cos^2(x) }}}


{{{1-cos^2(x)- 1=-cos^2(x)}}}


{{{-cos^2(x)=-cos^2(x)}}} ........{{{true}}}