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If A = 10°, find the value of \(\frac{12sin3A+5cos(5A-5)°}{9sin\frac{9A}{2}-4cos(5A+10)°}\) |
\(\frac{6\sqrt {2}+5}{9-2\sqrt {5}}\) \(\frac{6\sqrt {2} + 7}{9 - 2\sqrt {2}}\) \(\frac{6\sqrt {2}+5}{9-2\sqrt {2}}\) \(\frac{9 + 2\sqrt {2}}{6\sqrt {2} - 5}\) |
\(\frac{6\sqrt {2}+5}{9-2\sqrt {2}}\) |
\(\frac{12sin30°+5cos(45°)}{9sin45°-4cos60°}\) = \(\frac{\frac{12}{2}+\frac{5}{\sqrt {2}}}{\frac{9}{\sqrt {2}}-\frac{4}{2}}\) = \(\frac{12\sqrt {2}+10}{18-4\sqrt {2}}\) = \(\frac{2(6\sqrt {2}+5)}{2(9-2\sqrt {2})}\) =\(\frac{6\sqrt {2}+5}{9-2\sqrt {2}}\) |
