lim-x-0-tan-2x-pi-4-2tan-x-pi-4-tan-pi-4-sin-2x-pi-4-2sin-x-pi-4-sin-pi-4- Tinku Tara June 3, 2023 Limits 0 Comments FacebookTweetPin Question Number 132240 by bemath last updated on 12/Feb/21 limx→0tan(2x+π4)−2tan(x+π4)+tanπ4sin(2x+π4)−2sin(x+π4)+sinπ4? Answered by EDWIN88 last updated on 13/Feb/21 limx→02sec2(2x+π4)−2sec2(x+π4)2cos(2x+π4)−2cos(x+π4)=limx→0cos2(x+π4)−cos2(2x+π4)[cos(2x+π4).cos(x+π4)]2[cos(2x+π4)−cos(x+π4)]=limx→0−[cos(x+π4)+cos(2x+π4)][cos(2x+π4).cos(x+π4)]2=−2(12)2=−42 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: I-3x-5-x-2-2x-3-2-dx-Next Next post: What-is-the-set-Z-8-0-I-met-this-notation-in-a-question-asking-whether-or-not-the-set-Z-8-0-forms-a-group-under-multiplication-mod-8- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.