MCQ
The value of $\frac{\cos3\text{x}}{2\cos2\text{x}-1}$ is equal to:
- A$\cos\text{x}$
- B$\sin\text{x}$
- C$\tan\text{x}$
- DNone of these
✓
Answer
- $\cos\text{x}$
Solution:
We have,
$\therefore\frac{\cos3\text{x}}{2\cos2\text{x}-1}=\frac{4\cos^3\text{x}-3\cos\text{x}}{1(2\cos^2\text{x}-1)-1}$ $[\therefore\cos3\text{x}=4\cos^3\text{x}-3\cos\text{x}]$
$=\frac{4\cos^3\text{x}-3\cos\text{x}}{4\cos^2\text{x}-2-1}$
$=\frac{4\cos^3\text{x}-3\cos\text{x}}{4\cos^2\text{x}-3}$
$=\cos\text{x}\Big(\frac{4\cos^2\text{x}-3}{4\cos^2\text{x}-3}\Big)$
$=\cos\text{x}$
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