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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If 2x + 2cos x = 1, then ( ^ 2 x) (2 - ^ 2 x) is equal to

If cos⁡2x+2cosx=1,then\cos 2x + 2cos x = 1, then\left(\sin^{2} x\right)(2−cos⁡2x)\left(2 - \cos^{2} x\right) is equal to

Correct answer

1

Step-by-step solution

cos⁡2x+2cosx=1\cos 2x + 2cos x = 1
2cos⁡2x−1+2cosx=12\cos^{2} x - 1 + 2cos x = 1
cos⁡2x+cosx−1=0\cos^{2} x + cos x - 1 = 0
cosx=−1±52cos x = \dfrac{- 1 \pm \sqrt{5}}{2}
cosx=5−12⇒cos⁡2x=5+1−254=3−52cos x = \dfrac{\sqrt{5} - 1}{2} ⇒ \cos^{2} x = \dfrac{5 + 1 - 2\sqrt{5}}{4} = \dfrac{3 - \sqrt{5}}{2}
sin⁡2x(2−cos⁡2x)\sin^{2} x\left(2 - \cos^{2} x\right)
=(1−cos⁡2x)(2−cos⁡2x)= \left(1 - \cos^{2} x\right)\left(2 - \cos^{2} x\right)
=(1−(3−52))(2−(3−52))= \left(1 - \left(\dfrac{3 - \sqrt{5}}{2}\right)\right)\left(2 - \left(\dfrac{3 - \sqrt{5}}{2}\right)\right)
=(2−3+5)2(4−3+5)2= \dfrac{\left(2 - 3 + \sqrt{5}\right)}{2}\dfrac{\left(4 - 3 + \sqrt{5}\right)}{2}
=(5−1)2(5+1)2=44=1= \dfrac{\left(\sqrt{5} - 1\right)}{2}\dfrac{\left(\sqrt{5} + 1\right)}{2} = \dfrac{4}{4} = 1
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