Question

$\frac{tan41°}{cos49°}$⁰

Solve ?¿

Answer 1

$\sf\dfrac{tan 41°}{cos 49°}$

= $\sf\dfrac{tan (90-49)°}{cos 49°}$

= $\sf\dfrac{cot 49°}{cos 49°}$

= (cos 49°/sin49°)/cos 49°

= $\bold{\dfrac{1}{sin 49°}}$

Hope Helps uh..! :)

Answer 2

$\frac{ \tan(41)}{ \cos(49) } \\ = \frac{ \tan(90 - 49) }{ \cos(49) } \\ = \frac{ \cos(49) }( \cos(49) )} \\ \frac{1}{ \sin(45) }$