Test Index

HCU MCA 2013 Paper

© examsnet.com

Question : 73 of 75

Marks: +1, -0

If

\frac{2 \sin α}{1 + \cos α + \sin α} = y

\frac{1 - \cos α + \sin α}{1 + \sin α}

is equal to

$\frac{1}{y}$
1 - y
y
1+y

Solution:

(c) Given that,

\;\;\;\;\;\;\;\;\;\;\; \frac{2 \sin α}{1 + \cos α + \sin α} = y

⇒

y = \frac{2 \sin α}{1 + 2 \cos^2 \frac{α}{2} - 1 + 2 \sin \frac{α}{2} \cdot \cos \frac{α}{2}}

\Rightarrow y = \frac{2 \times 2 \sin \frac{α}{2} \times \cos \frac{α}{2}}{2 \cos \frac{α}{2} \left(\cos \frac{α}{2} + \sin \frac{α}{2}\right)}

\Rightarrow y = \frac{2 \sin \frac{α}{2}}{\sin \frac{α}{2} + \cos \frac{α}{2}} = \frac{2 \tan \frac{α}{2}}{1 + \tan \frac{α}{2}} \;\;\;\;\;\;\;\;\;\;...(i)

\Rightarrow y + y \tan \frac{α}{2} = 2 \tan \frac{α}{2}

\Rightarrow y = (2 - y) \tan \frac{α}{2}

\Rightarrow \tan \frac{α}{2} = \frac{y}{2 - y} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(ii)

Now, we have

\frac{1 - \cos α + \sin α}{1 + \sin α} = \frac{1 - \left(1 - 2 \sin^2 \frac{α}{2}\right) + 2 \sin \frac{α}{2} \cdot \cos \frac{α}{2}}{1 + \sin α}

= \frac{2 \sin \frac{α}{2} \left(\sin \frac{α}{2} + \cos \frac{α}{2}\right)}{\left(\sin^2 \frac{α}{2} + \cos^2 \frac{α}{2}\right) + 2 \sin \frac{α}{2} \cdot \cos \frac{α}{2}}

= \frac{2 \sin \frac{α}{2} \left(\sin \frac{α}{2} + \cos \frac{α}{2}\right)}{\left(\sin \frac{α}{2} + \cos \frac{α}{2}\right)^2} = \frac{2 \sin \frac{α}{2}}{\sin \frac{α}{2} + \cos \frac{α}{2}}

=

y \;\;\;\;\;\;\;\;\;\;\;\;\;\;[\text{from Eq. (i)}]

© examsnet.com

Go to Question:

Prev Question

Next Question