If sin x = 12/13, where 0° < x < 90°, find the value of 1 - cos\(^2\)x

  • A 25/169
  • B 64/169
  • C 105/169
  • D 144/169
  • E 8/13
waec 1993mathematics

The correct answer is D. 144/169

\(\sin x = \frac{12}{13}\)

\(\cos x = \frac{5}{13}\)

\(\cos^{2} x = (\frac{5}{13})^2 = \frac{25}{169}\)

\(1 - \cos^{2} x = 1 - \frac{25}{169}\)

= \(\frac{144}{169}\)

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