Question

$\frac{sin\Theta}{1+cos\Theta}$ is equal to
(a)$\frac{1+cos\Theta}{sin\Theta}$
(b)$\frac{1-cos\Theta}{sin\Theta}$
(c)$\frac{1-cos\Theta}{cos\Theta}$
(d)$\frac{1-sin\Theta}{cos\Theta}$

Answer 1

Answer:

sinθ/(1 + cosθ) is equal to  (1 - cosθ)/(sinθ).

Among the given options option (b)  (1 - cosθ)/(sinθ) is correct.  

Step-by-step explanation:

Given : sinθ/(1 + cosθ)

= sinθ × (1 - cosθ)/[(1 + cosθ)(1 - cosθ)]

[Multiplying the numerator and denominator by (1 - cosθ)]

= sinθ × (1 - cosθ)/(1 - cos²θ)

[By using an identity , (a + b) (a - b) = a² - b²]

= sinθ × (1 - cosθ)/(sin²θ)

[By using  an identity, (1- cos²θ) = sin²θ]

sinθ/(1 + cosθ)  = (1 - cosθ)/(sinθ)

Hence, sinθ/(1 + cosθ) is equal to  (1 - cosθ)/(sinθ).

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

$\displaystyle{\implies\left(\dfrac{1-\cos \theta}{\sin \theta}\right)}$

Option b.  is correct .

Step-by-step explanation:

Given :

$\displaystyle{\dfrac{\sin \theta}{1+\cos \theta}}$

Multiplying and dividing by 1 - cos θ

$\displaystyle{\dfrac{\sin \theta}{1+\cos \theta}\times\dfrac{1-\cos \theta}{1-\cos \theta}}\\\\\\\displaystyle{\dfrac{\sin \theta\left(1-\cos \theta)}{1-\cos^2 \theta}}$

Now using identity

$\displaystyle{1-\cos^2 \theta = \sin^2\theta }$

$\displaystyle{\implies\dfrac{\sin \theta\left(1-\cos \theta)}{\sin^2 \theta}}\\\\\\\displaystyle{\implies\dfrac{\sin \theta\left(1-\cos \theta)}{\sin \theta\times\sin \theta}}\\\\\\\displaystyle{\implies\left(\dfrac{1-\cos \theta}{\sin \theta}\right)}$

Thus we get answer .