Question

Anyone plzz solve it fast....​

Answer 1

Answer:

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

$= > \frac{1 + \cos(0) - {sin}^{2} (0)}{sin(0) \: (1 + cos(0))} = cot(0)$

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:-

$\tt {Additional \: Information}$

$= > { \cos }^{2} (0) + {sin}^{2} (0) = 1$

$= > 1 + {tan}^{2} (0) = {sec}^{2} (0)$

$= > {cot}^{2} (0) + 1 = {cosec}^{2} (0)$

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