Question

pls ans this question we have to evaluate ​

Answer 1

$\frac{ \sin(35) \cos(55) + \cos(35) \sin(55) }{ {cosec}^{2} 10 - {tan}^{2} 80} \\ = \frac{ \sin(90 - 55) \cos(55) + \cos(90 - 55) \sin(55) }{ {cosec}^{2} 10 - {tan}^{2} 80} \\ = \frac{ \cos(55) \cos(55) + \sin(55) \sin(55) }{ {cosec}^{2} 10 - {tan}^{2} 80} \\ = \frac{ \ \ {cos}^{2} (55) + \ {sin}^{2} (55) }{ 1 + {cot}^{2} 10 - {tan}^{2} 80} \\ = {cos}^{2} (55) + \ {sin}^{2} (55) }{ 1 + 1} \\ = \frac{1}{2}$