Question

0%5Ctheta%29%20%7D%20%20%5C%3A%20is%20%5C%3A%20equal%20%5C%3A%20to%20%5C%5C%20" id="TexFormula1" title="9\: \: \frac{ \sin( \theta) }{1 + \cos( \theta) } \: is \: equal \: to \\ " alt="9\: \: \frac{ \sin( \theta) }{1 + \cos( \theta) } \: is \: equal \: to \\ " align="absmiddle" class="latex-formula">
$(a) \: \frac{1 + \cos( \theta) }{ \sin( \theta) } \\ (b) \frac{1 - \cos( \theta) }{ \sin( \theta) } \\ (c) \: \frac{1 - \cos( \theta) }{ \cos( \theta) } \\ (d) \: \frac{1 - \sin( \theta) }{ \cos( \theta) }$
Correct Explanation with be mark as brainliest.​

Answer 1

Answer:

need explaintion pleaawee

Answer 2

Option (b) is your correct answer.

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{sin\theta }{1 + cos\theta } \times \dfrac{1 - cos\theta }{1 - cos\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using this, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{1 - {cos}^{2}\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{{sin}^{2}\theta }$

$\rm \: = \: \dfrac{1 - cos\theta }{sin\theta }$

Hence,

$\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta } = \: \dfrac{1 - cos\theta }{sin\theta } }}}$