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PilotLPTM [1.2K]
2 years ago
12

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.​
Mathematics
2 answers:
LekaFEV [45]2 years ago
7 0

Answer:

need explaintion pleaawee

PIT_PIT [208]2 years ago
3 0

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 }

<u>Hence, </u>

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

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