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NDA 1 2025 Maths, question 35

Topic: Trigonometry

Passage · Q34 to Q35

Let p=∣sin⁡α−sin⁡(α−90∘)∣p = \left| \sin\alpha - \sin\left( \alpha - 90^{\circ} \right) \right|.

Q35

What is the maximum value of pp?

  1. (a) 1
  2. (b) 2\sqrt{2}Correct answer
  3. (c) 3\sqrt{3}
  4. (d) 2

Solution.

Answer: (b) 2\sqrt{2}

  1. Using p=2∣sin⁡(α+45∘)∣p = \sqrt{2}\left| \sin\left( \alpha + 45^{\circ} \right) \right| and ∣sin⁡t∣≤1\left| \sin t \right| \leq 1 gives p≤2p \leq \sqrt{2}.
  2. Equality occurs at α=45∘\alpha = 45^{\circ}, so the maximum is 2\sqrt{2}.