MCQ
The value of $\frac{\cos3\text{x}}{2\cos2\text{x}-1}$ is equal to:
  • $\cos\text{x}$
  • B
    $\sin\text{x}$
  • C
    $\tan\text{x}$
  • D
    None of these

Answer

Correct option: A.
$\cos\text{x}$
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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