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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2025 · Mathematics · Trigonometric Ratios & Identities

Consider the following for the two (02) items that follow: Let p =...

Consider the following for the two (02) items that follow:
Let p=∣sin⁡α−sin⁡(α−90°)∣p = |\sin\alpha - \sin(\alpha - 90°)|.

What is the maximum value of pp?

Options

  1. A11
  2. B2\sqrt{2}
  3. C3\sqrt{3}
  4. D22

Correct answer

B. 2\sqrt{2}

Step-by-step solution

Given p = |\sin\alpha - \sin(\alpha - 90^{}^{\circ})|

Using the trigonometric identity \sin(\alpha - 90^{}^{\circ}) = -\sin(90^{}^{\circ} - \alpha) = -\cos\alpha

Substituting this into the expression for pp

p=∣sin⁡α−(−cos⁡α)∣=∣sin⁡α+cos⁡α∣p = |\sin\alpha - (-\cos\alpha)| = |\sin\alpha + \cos\alpha|

The maximum value of the expression asin⁡α+bcos⁡αa\sin\alpha + b\cos\alpha is a2+b2<br><br>Here,\sqrt{a^2 + b^2}<br><br>Here,a = 1andandb = 1<br><br>Maximumvalueofsin⁡α+cos⁡α=12+12=2<br><br>Maximum value of \sin\alpha + \cos\alpha = \sqrt{1^2 + 1^2} = \sqrt{2}

Therefore, the maximum value of pp is 2<br><br>Answer:2\sqrt{2}<br><br>Answer: \sqrt{2}
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