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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
The value of $\frac{2(\sin2\text{x}+2\cos^2\text{x}-1)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$ is:
  • A
    $\cos\text{x}$
  • B
    $\sec\text{x}$
  • ✓
    $\text{cosec}\ \text{x}$
  • D
    $\sin\text{x}$
✓

Answer

Correct option: C.
$\text{cosec}\ \text{x}$
We have,
$\frac{2(\sin2​\text{x}+2\cos^2\text{x}-1​)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{\cos\text{x}-\sin\text{x}-4\cos^3\text{x}+3\cos\text{x}\sin\text{x}-4\sin^3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}-4\cos^3\text{x}+2\sin\text{x}-4\sin^3\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}(1-\cos^2​​​​\text{x})+2\sin\text{x}(1-2\sin^2\text{x})}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}\sin^2\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\times2\sin\text{x}\cos\text{x}\sin\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin2\text{x}\sin2\text{x}+2\sin\text{x}\cos2\text{x}}$
$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin\text{x}(\sin2\text{x}+\cos2\text{x})}$
$=\frac{1}{\sin\text{x}}$
$=\text{cosec}\ \text{x}$

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