if x = 2sin(theta)/(1 + cos(theta) + sin(theta)), then 1-cos(theta)+ sin(theta)/(1+sin(theta) =  a) 1+x  b)1-x   c) x d)1/x

Dear Student,

$⇒$$\frac{1-\mathrm{cos\theta }+\mathrm{sin\theta }}{1+\mathrm{sin\theta }}$ = $\frac{1-\mathrm{cos\theta }+\mathrm{sin\theta }}{1+\mathrm{sin\theta }}$$\frac{1+\mathrm{cos\theta }+\mathrm{sin\theta }}{1+\mathrm{cos\theta }+\mathrm{sin\theta }}$
= $\frac{\left[\left(1+\mathrm{sin}\theta \right)-\mathrm{cos}\theta \right]\left[\left(1+\mathrm{sin\theta }\right)+\mathrm{cos}\theta \right]}{\left(1+sin\theta \right)\left(1+\mathrm{sin\theta }+\mathrm{cos\theta }\right)}$
=          (Using (a-b)(a+b) = a2 - b2 )
=
=       (Using sin2A + cos2A = 1 )
= $\frac{\left[\left(2si{n}^{2}\theta +2\mathrm{sin}\theta \right)\right]}{\left({\left(1+\mathrm{sin}\theta \right)}^{2}+cos\theta \left(1+sin\theta \right)\right)}$
= $\frac{2sin\theta \left(sin\theta +1\right)}{\left(1+sin\theta \right)\left[\left(1+\mathrm{sin}\theta \right)+\mathrm{cos}\theta \right]}$
= $\frac{2sin\theta }{\left[\left(1+\mathrm{sin}\theta \right)+\mathrm{cos}\theta \right]}$
=   x                                              (As per given expression for x)

Hence option c is correct

Regards

• 28
Hello Abhiram Rangan, option C ie x is the correct one
• -17
How??
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