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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2009 · Mathematics · Trigonometric Equations

If (y= (2 α)/(1+ α+ α) ), then value of ( (1- α+ α)/(1+ α) ) is

If (y=2sin⁡α1+cos⁡α+sin⁡α\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}), then value of (1−cos⁡α+sin⁡α1+sin⁡α\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}) is

Options

  1. A(y3\frac{y}{3})
  2. B(y\mathrm{y})
  3. C2y2 y
  4. D(32\frac{3}{2}y)

Correct answer

B. (y\mathrm{y})

Step-by-step solution

1−cos⁡α+sin⁡α1+sin⁡α=1−cos⁡α+sin⁡α1+sin⁡α⋅1+cos⁡α+sin⁡α1+cos⁡α+sin⁡α=(1+sin⁡α)2−cos⁡2α(1+sin⁡α)(1+cos⁡α+sin⁡α)=2sin⁡α(1+sin⁡α)(1+sin⁡α)(1+cos⁡α+sin⁡α)=2sin⁡α1+cos⁡α+sin⁡α=y\begin{aligned} \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}\\ & =\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}\\ & =\frac{(1+\sin \alpha)^2-\cos ^2 \alpha}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}\\ & =\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\cos \alpha+\sin \alpha)}\\ & =\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y \end{aligned})
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