Question

Solve this.........​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Step-by-step explanation:

$\frac{sin\theta}{1 + cos\theta} + \frac{1 + cos\theta}{sin\theta} \\ = > \frac{( sin\theta \: sin\theta \: + \: (1 + cos\theta) (1 + cos\theta))}{(1 + cos\theta) \: sin\theta} \\ = > \frac{ {sin}^{2}\theta \: + \: 1 + 2cos\theta \: + {cos}^{2} \theta}{ \: \: (1 + cos\theta) \: sin\theta} \\ = > \frac{( {sin}^{2}\theta\: + {cos}^{2} \theta) + 1 + 2cos\theta}{ \: (1 + cos\theta) \: sin\theta } \\ = > \frac{( \: \: \ 1+ 1 + 2cos\theta \: )}{ \: \: (1 + cos\theta) \: sin\theta\: \: } \\ = > \frac{2(1 + cos\theta)}{ \: (1 + cos\theta) \: sin\theta \: \: } \\ = > \frac{2}{sin\theta} \\ = > 2cosec\theta$