If A = 10º, what is the value of : $\frac{12sin3A+5cos(5A-5^o)}{9sin\frac{9A}{2}-4cos(5A+10^o)}$?
Answer & explanation
Correct answer: option 1
$\frac{12sin3A+5cos(5A-5^o)}{9sin\frac{9A}{2}-4cos(5A+10^o)}$
= \(\frac{12 sin30º 5cos(50º - 5º)}{9sin45º - 4 cos 60º }\)
= \(\frac{12 × 1/2 + 5/√2}{9× 1/√2 - 4 × 1/2 }\)
= \(\frac{6√2 + 5}{9- 2√2 }\)