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Mathematics · JEE | TRIGONOMETRIC FUNCTIONS

हिन्दी

The value of \( \cot^2 x + \csc^2 x \) when \( x = \frac{\pi}{4} \) is:

  1. A
    \( 1 \)
  2. B
    \( 2 \)
  3. C
    \( 3 \)
  4. D
    \( 4 \)
View Answer & Solution

Correct answer · Option D

\( 4 \)

Explanation

At \( x = \frac{\pi}{4} \), \( \cot(\frac{\pi}{4}) = 1 \) and \( \csc(\frac{\pi}{4}) = \sqrt{2} \). Thus, \( \cot^2(\frac{\pi}{4}) + \csc^2(\frac{\pi}{4}) = 1^2 + (\sqrt{2})^2 = 1 + 2 = 3 \) — wait, let's re-verify: \( 1 + 2 = 3 \). Option 3 is correct.
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