Question

Alguien me ayuda con identidades trigonométricas? cotB+tanB=secBcscB
Por favor..!!

Answer 1

Recordemos:
• tanx = senx/cosx
• ctgx = cosx/senx
• sen²x + cos²x = 1

El ejercicio:
${\cot x +\tan x= \sec x.\csc x}\\\\{\cot x +\tan x=}\\\\\\{\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}=}\\\\\\{\dfrac{\cos^2x+\sin^2x}{\sin x.\cos x}=}\\\\\\{\dfrac{1}{\sin x.\cos x}=}\\\\\\{\dfrac{1}{\sin x}.\dfrac{1}{\cos x}=}\\\\\\{\boxed{\csc x. \sec x}}$


Salu2.!! :)
Wellington