Option (b) is your correct answer.
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
Given Trigonometric expression is
[tex]\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }[/tex]
So, on rationalizing the denominator, we get
[tex]\rm \: = \: \dfrac{sin\theta }{1 + cos\theta } \times \dfrac{1 - cos\theta }{1 - cos\theta } [/tex]
We know,
[tex] \purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}[/tex]
So, using this, we get
[tex]\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{1 - {cos}^{2}\theta } [/tex]
We know,
[tex] \purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}[/tex]
So, using this identity, we get
[tex]\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{{sin}^{2}\theta } [/tex]
[tex]\rm \: = \: \dfrac{1 - cos\theta }{sin\theta } [/tex]
Hence,
[tex] \\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta } = \: \dfrac{1 - cos\theta }{sin\theta } }}} \\ [/tex]
