JAMB - Mathematics (1984 - No. 34)
If sin \(\theta\) = \(\frac{x}{y}\) and 0o < 90o then find \(\frac{1}{tan\theta}\)
\(\frac{x}{\sqrt{y^2 - x^2}}\)
\(\frac{y}{x}\)
\(\frac{\sqrt{y^2 + x^2}}{y^2 - x^2}\)
\(\frac{y^2 - x^2}{x}\)
Giải thích
\(\frac{1}{tan\theta}\) = \(\frac{cos\theta}{sin\theta}\)
sin\(\theta\) = \(\frac{x}{y}\)
cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)
sin\(\theta\) = \(\frac{x}{y}\)
cos\(\theta\) = \(\frac{\sqrt{y^2 - x^2}}{y}\)