Question
$\frac{\sin3\text{x}}{1+2\cos2\text{x}}$ is equal to:
- $\cos\text{x}$
- $\sin\text{x}$
- $-\cos\text{x}$
- $\sin\text{x}$
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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$\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ is:
$\frac{1}{10}$
$-\frac{1}{10}$
$1$
None of these.