Question 1079875


{{{sin^2 (x) cot^2 (x )}}}...............since {{{cot^2(x) = (1/(tan(x)))^2}}}, we have

={{{sin^2 (x) (1/(tan(x)))^2}}}............and {{{tan(x)=sin(x)/cos(x)}}}, so we have


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


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


={{{sin^2 (x)(cos^2(x)/sin^2(x))}}}................cancel {{{sin^2 (x) }}}


={{{cos^2(x)}}}...........since {{{cos^2(x)=1-sin^2(x)}}} , we have


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


answer: c. {{{1 - sin^2(x)}}}