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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2015 · Mathematics · Trigonometric Ratios & Identities

(1- A+ A)^2 is equal to

(1−sin⁡A+cos⁡A)2(1-\sin A+\cos A)^2 is equal to

Options

  1. A2(1−cos⁡A)(1+sin⁡A)2(1-\cos A)(1+\sin A)
  2. B2(1−sin⁡A)(1+cos⁡A)2(1-\sin A)(1+\cos A)
  3. C2(1−cos⁡A)(1−sin⁡A)2(1-\cos A)(1-\sin A)
  4. DNone of the above

Correct answer

B. 2(1−sin⁡A)(1+cos⁡A)2(1-\sin A)(1+\cos A)

Step-by-step solution

(1−sin⁡A+cos⁡A)2=1+sin⁡2A+cos⁡2A−2sin⁡A−2sin⁡A⋅cos⁡A+2cos⁡A\begin{aligned} & (1-\sin A+\cos A)^2 \\ & =1+\sin ^2 A+\cos ^2 A-2 \sin A \\ & \quad-2 \sin A \cdot \cos A+2 \cos A\end{aligned}=2−2sin⁡A−2sin⁡Acos⁡A+2cos⁡A=2(1−sin⁡A)+2cos⁡A(1−sin⁡A)=2(1+cos⁡A)(1−sin⁡A)\begin{aligned} & =2-2 \sin A-2 \sin A \cos A+2 \cos A \\ & =2(1-\sin A)+2 \cos A(1-\sin A) \\ & =2(1+\cos A)(1-\sin A) \end{aligned}\therefore$ Option (B) is correct.
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