If A = 10°, find the value of \(\frac{12sin3A+5cos(5A-5)°}{9sin\frac{9A}{2}-4cos(5A+10)°}\)
Answer & explanation
Correct answer: option 3
\(\frac{12sin30°+5cos(45°)}{9sin45°-4cos60°}\) = \(\frac{\frac{12}{2}+\frac{5}{\sqrt {2}}}{\frac{9}{\sqrt {2}}-\frac{4}{2}}\)
= \(\frac{12\sqrt {2}+10}{18-4\sqrt {2}}\)
= \(\frac{2(6\sqrt {2}+5)}{2(9-2\sqrt {2})}\)
=\(\frac{6\sqrt {2}+5}{9-2\sqrt {2}}\)