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Sin 3 X 1 + 2 Cos 2 X is Equal to

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प्रश्न

\[\frac{\sin 3x}{1 + 2 \cos 2x}\]   is equal to

विकल्प

  • cos x

  • sin x

  •  – cos x

  • sin x

MCQ
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उत्तर

sin 

\[\text{ We have } , \]

\[\frac{\sin 3x}{1 + 2\cos 2x} = \frac{3\text{ sin } x - 4 \sin^3 x}{1 + 2\left( 1 - 2 \sin^2 x \right)}\]

\[ = \frac{3\text{ sin } x - 4 \sin^3 x}{1 + 2 - 4 \sin^2 x}\]

\[ = \frac{\text{ sin } x\left( 3 - 4 \sin^2 x \right)}{\left( 3 - 4 \sin^2 x \right)}\]

\[ = \sin x\]

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [पृष्ठ ४४]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 21 | पृष्ठ ४४

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