Question

121 sum solution.
please answer.

Answer 1

easy dude.........practice more...........

Answer 2

121.

$= \ \textgreater \ Given : \frac{cotA+ tanB}{cotB + tanA}$

$= \ \textgreater \ \frac{ \frac{cosA}{sinA} + \frac{sinB}{cosB} }{ \frac{cosB}{sinB} + \frac{sinA}{cosA} }$


$= \ \textgreater \ \frac{ \frac{cosAcosB + sinAsinB}{sinAcosB} }{ \frac{cosAcosB + sinAsinB }{cosAsinB} }$


$= \ \textgreater \ \frac{cosAsinB(cosAcosB +sinAsinB) }{sinAcosB(cosAcosB + sinAsinB)}$


$= \ \textgreater \ \frac{cosAsinB}{sinAcosB}$

$= \ \textgreater \ \frac{cosA}{sinA} * \frac{sinB}{cosB}$

= > cotA . tanB




Hope this helps!