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प्रश्न
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
विकल्प
cos x
sec x
cosec x
sin x
उत्तर
cosec x
\[\text{ We have } , \]
\[\frac{2\left( \sin2x + 2 \cos^2 x - 1 \right)}{\text{ cos } x - \text{ sin } x - \cos3x + \sin3x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{\text{ cos } x - \text{ sin } x - 4 \cos^3 x + 3\text{ cos } x + 3\text{ sin } x - 4 \sin^3 x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{4\text{ cos } x - 4 \cos^3 x + 2\text{ sin } x - 4 \sin^3 x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{4\text{ cos } x\left( 1 - \cos^2 x \right) + 2\text{ sin } x\left( 1 - 2 \sin^2 x \right)}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{4\text{ cos } x \sin^2 x + 2\text{ sin } x \cos2x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{2 \times 2\text{ sin } x \text{ cos } x \text{ sin } x + 2\text{ sin } x \cos2x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{2\sin2x \text{ sin } x + 2\text{ sin } x \cos2x}\]
\[ = \frac{2\left( \sin2x + \cos2x \right)}{2\text{ sinx } \left( \sin2x + \cos2x \right)}\]
\[ = \frac{1}{\text{ sin } x}\]
\[ = \text{ cosec } x \]
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