KeerthanaPosted on sec4θ−sec2θsec ^{4} heta-sec ^{2} hetasec4θ−sec2θ is equal to atan2 T – tan4 T bcos2 T – cos4 T ctan2 T + tan4 T dcos4 T – cos2 T Answer : Option CExplanation : sec4θ−sec2θsec ^{4} heta-sec ^{2} hetasec4θ−sec2θ =sec2θ(sec2θ−1)=sec ^{2} hetaleft(sec ^{2} heta-1 ight)=sec2θ(sec2θ−1) =(1+tan2θ)(1+tan2θ−1)=left(1+ an ^{2} heta ight)left(1+ an ^{2} heta-1 ight)=(1+tan2θ)(1+tan2θ−1) =tan2θ+tan4θ= an ^{2} heta+ an ^{4} heta=tan2θ+tan4θ Rate This:NaN / 5 - 1 votesAdd comment
