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If Cos 2 X + 2 Cos X = 1 Then, ( 2 − Cos 2 X ) Sin 2 X is Equal to

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प्रश्न

If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

 
 

विकल्प

  • 1

  • -1

  • \[- \sqrt{5}\]

     

  • \[\sqrt{5}\]

     

MCQ
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उत्तर

1

\[We have, \]
\[\cos2x + 2\text{ cos } x = 1\]
\[ \Rightarrow 2 \cos^2 x - 1 + 2\text{ cos } x = 1\]
\[ \Rightarrow \cos^2 x + \text{ cos } x - 1 = 0\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 \pm \sqrt{1^2 + 4}}{2}\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 \pm \sqrt{5}}{2}\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 + \sqrt{5}}{2}\]

\[\text{ Now, }  \]
\[\left( 2 - \cos^2 x \right) \sin^2 x = \left[ 2 - \left( \frac{- 1 + \sqrt{5}}{2} \right)^2 \right] \left( 1 - \cos^2 x \right)\]
\[ = \left[ 2 - \frac{1}{4}\left( 1 - 2\sqrt{5} + 5 \right) \right] \left( 1 - \frac{1}{4}\left( 1 - 2\sqrt{5} + 5 \right) \right)\]
\[ = \frac{1}{4}\left( 1 + \sqrt{5} \right)\left( \sqrt{5} - 1 \right) = \frac{4}{4} = 1\]

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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 ४३]

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 4 | पृष्ठ ४३

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