UPSC NDA I 2025 Mathematics Paper

Given,
The expression for p is:
p=|sin‌α−sin‌(α−90∘)|
Using the identity for sin‌(α−90∘)
sin‌(α−90∘)=cos‌α
Substituting this into the expression for p :
p=|sin‌α−cos‌α|
The expression |sin‌α−cos‌α| achieves its maximum value when the difference between sin‌α and cos‌α is largest.
We can rewrite sin‌α−cos‌α as:
sin‌α−cos‌α=√‌2(sin‌(α−45∘))
Since sin‌(α−45∘) has a maximum value of 1 , the maximum value of √‌2(sin‌(α−45∘)) is:
√‌2