Anyone plzz solve it fast....

1.	Anyone plzz solve it fast....
Answer» ANSWER: $\bold {According \: to \: Question }$ $= > \frac{<klux>1</klux> + \cos(<klux>0</klux>) - {sin}^{2} (0)}{sin(0) \: (1 + cos(0))} = cot(0)$ *We can* also write as:-** $\frac{1 + \cos(0) - {sin}^{2} (0) + {cos}^{2}(0) - {cos}^{2} (0) }{sin(0) \: (1 + cos(0))} = cot(0)$ here, $- ({sin}^{2} (0) - {cos}^{2} (0) )= - 1$ Therefore, it becomes:- $\frac{1 - 1 + cos(0) + {cos}^{2}(0) }{sin(0) \: (1 + cos(0))} = cot(0)$ $\frac{ cos(0) + {cos}^{2}(0) }{sin(0) \: (1 + cos(0))} = cot(0)$ $= > \frac{cos(0) \: (1 + cos(0))}{sin(0) \: (1 + cos(0))} = cot(0)$ Cancelling 1+ cos(0) by 1+ cos(0), we get:- $= > \frac{cos(0)}{ \sin(0) } = \cot(0)$ $= > \cos(0) = \cot(0)$ *Hence* *proved*:- $\tt {Additional \: Information}$ $= > { \cos }^{2} (0) + {sin}^{2} (0) = 1$ $= > 1 + {tan}^{2} (0) = {sec}^{2} (0)$ $= > {cot}^{2} (0) + 1 = {cosec}^{2} (0)$ hope it helps you... FOLLOW ME...

Answer»

$\bold {According \: to \: Question }$