WAEC - Mathematics (1993 - No. 32)
If sin x = 12/13, where 0° < x < 90°, find the value of 1 - cos\(^2\)x
25/169
64/169
105/169
144/169
8/13
Explanation
\(\sin x = \frac{12}{13}\)
\(\cos x = \frac{5}{13}\)
\(\cos^{2} x = (\frac{5}{13})^2 = \frac{25}{169}\)
\(1 - \cos^{2} x = 1 - \frac{25}{169}\)
= \(\frac{144}{169}\)
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