Question

The value of $\frac{\tan {41}^{\circ }}{\cos {49}^{\circ }}$ = $\text{cosec}\theta$, then $\theta$ =

Answer 1

$\frac{{\tan 41}^0}{{\cos 49}^0}=\frac{\tan ({90}^0-{49}^0)}{\cos {49}^0}=\frac{{\cot 49}^0}{\cos {49}^0}$

$(\because \tan ({90}^0-\theta )=\cot \theta )$

$=\frac{{\cos 49}^0}{{\sin 49}^0.{\cos 49}^0}$

$(\because \cot \theta =\frac{\cos \theta }{\sin \theta })$

$=\frac{1}{{\sin 49}^0}$

$=cosec{49}^0$