MCQOPTIONS
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| 1. |
If sec θ – tan θ = x/y, (0 < x < y) and 0°< θ < 90°, then sinθ is equal to: |
| A. | \(\frac{{{y^2} - {x^2}}}{{{x^2} + {y^2}}}\) |
| B. | \(\frac{{{x^2} + {y^2}}}{{{y^2} - {x^2}}}\) |
| C. | \(\frac{{2xy}}{{{x^2} + {y^2}}}\) |
| D. | \(\frac{{{x^2} + {y^2}}}{{2xy}}\) |
| Answer» B. \(\frac{{{x^2} + {y^2}}}{{{y^2} - {x^2}}}\) | |