MCQ
$\frac{\sin3\text{x}}{1+2\cos2\text{x}}$ is equal to:
- A$\cos\text{x}$
- B$\sin\text{x}$
- C$-\cos\text{x}$
- D$\sin\text{x}$
Solution:
We have,
$\frac{\sin3\text{x}}{1+2\cos2\text{x}}=\frac{3\sin\text{x}-4\sin^2\text{x}}{1+2(1-2\sin^2\text{x})}$
$=\frac{3\sin\text{x}-4\sin^3\text{x}}{1+2-4\sin\text{x}}$
$=\frac{\sin\text{x}(3-4\sin^2\text{x})}{(3-4\sin^2\text{x})}$
$=\sin\text{x}$
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