Question Number 134902 by deleteduser12 last updated on 08/Mar/21
Commented by bobhans last updated on 08/Mar/21
$$\mathrm{use}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{sec}\:\mathrm{2x}\right)\mathrm{cot}\:\mathrm{2x}\:=\:\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{sin}\:\mathrm{2x}}\:=\mathrm{cot}\:\mathrm{x} \\ $$
Answered by bobhans last updated on 08/Mar/21
