The value of $\sec {41}^o\sin {49}^o+\cos {49}^o\text{cosec}{41}^o$ is :

The correct option is A $2$
$\sec {41}^o\sin {49}^o+\cos {49}^{o}cosec{41}^o$
$=\frac{\sin {49}^o}{\cos {41}^o}+\frac{\cos {49}^o}{\sin {41}^o}$
$=\frac{\sin {49}^o}{\cos ({90}^o-{49}^o)}+\frac{\cos {49}^o}{\sin ({90}^o-{49}^o)}$
$=\frac{\sin {49}^o}{\sin {49}^o}+\frac{\cos {49}^o}{\cos {49}^o}$
$=1+1=2$