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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

BITSAT 2015 · Mathematics · Inverse Trigonometric Functions

If ^ -1 x- ^ -1 (y)/(2)=α , then 4 x^ 2 -4 x y α+y^ 2 is equal to

If cos⁡−1x−cos⁡−1y2=α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha, then 4x2−4xycos⁡α+y24 x^{2}-4 x y \cos \alpha+y^{2} is equal to

Options

  1. A2sin⁡2α2 \sin 2 \alpha
  2. B4
  3. C4sin⁡2α4 \sin ^{2} \alpha
  4. D−4sin⁡2α-4 \sin ^{2} \alpha

Correct answer

C. 4sin⁡2α4 \sin ^{2} \alpha

Step-by-step solution

cos⁡−1x−cos⁡−1y2=α⇒cos⁡−1(xy2+(1−x2)(1−y24))=α⇒cos⁡−1(xy+4−y2−4x2+x2y22)=α⇒4x2+y2−4xycos⁡α=4sin⁡2α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha\Rightarrow \cos ^{-1}\left(\frac{x y}{2}+\sqrt{\left(1-x^{2}\right)\left(1-\frac{y^{2}}{4}\right)}\right)=\alpha\Rightarrow \cos ^{-1}\left(\frac{x y+\sqrt{4-y^{2}-4 x^{2}+x^{2} y^{2}}}{2}\right)=\alpha\Rightarrow 4 x^{2}+y^{2}-4 x y \cos \alpha=4 \sin ^{2} \alpha
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