NTA Abhyas 2020
Exam NTA Abhyas
Year 2020
Subject Mathematics
Chapter Trigonometric Ratios & Identities NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities
Let A = x ^ 4 + 4 x ^ 3 + 2 x ^ 2 - 4x + 7 where x = cot (11π)/(8)...
Let
A = x 4 + 4 x 3 + 2 x 2 − 4 x + 7 A = {x}^{4} + 4{x}^{3} + 2{x}^{2} - 4x + 7 A = x 4 + 4 x 3 + 2 x 2 − 4 x + 7 where
x = c o t 11 π 8 x = cot \dfrac{11\pi}{8} x = co t 8 11 π and
B = 1 − cos 8 θ tan 2 4 θ + 1 + cos 8 θ cot 2 4 θ B = \dfrac{1 - \cos 8\theta}{\tan^{2} 4\theta} + \dfrac{1 + \cos 8\theta}{\cot^{2} 4\theta} B = tan 2 4 θ 1 − cos 8 θ + cot 2 4 θ 1 + cos 8 θ where
θ = 9 o , \theta = {9}^{o}, θ = 9 o , then the value of
A × B 2 \dfrac{A \times B}{2} 2 A × B is equal to
Step-by-step solution
A = x 4 + 4 x 3 + 2 x 2 − 4 x + 7 A = {x}^{4} + 4{x}^{3} + 2{x}^{2} - 4x + 7 A = x 4 + 4 x 3 + 2 x 2 − 4 x + 7 where
x = c o t 11 π 8 = cot ( π + 3 π 8 ) = c o t 3 π 8 x = cot \dfrac{11\pi}{8} = \cot \left(\pi + \dfrac{3\pi}{8}\right) = cot \dfrac{3\pi}{8} x = co t 8 11 π = cot ( π + 8 3 π ) = co t 8 3 π = cot ( π 2 − π 8 ) = t a n π 8 = 2 − 1 = \cot \left(\dfrac{\pi}{2} - \dfrac{\pi}{8}\right) = tan \dfrac{\pi}{8} = \sqrt{2} - 1 = cot ( 2 π − 8 π ) = t an 8 π = 2 − 1 i.e.
x = 2 − 1 ⇒ x 2 = 2 + 1 − 2 2 = 1 − 2 ( 2 − 1 ) x = \sqrt{2} - 1 ⇒ {x}^{2} = 2 + 1 - 2\sqrt{2} = 1 - 2\left(\sqrt{2} - 1\right) x = 2 − 1 ⇒ x 2 = 2 + 1 − 2 2 = 1 − 2 ( 2 − 1 ) ⇒ x 2 = 1 − 2 x ⇒ x 2 + 2 x − 1 = 0 ⇒ {x}^{2} = 1 - 2x ⇒ {x}^{2} + 2x - 1 = 0 ⇒ x 2 = 1 − 2 x ⇒ x 2 + 2 x − 1 = 0 So,
A = ( x 2 + 2 x − 1 ) 2 + 6 = 6 A = {\left({x}^{2} + 2x - 1\right)}^{2} + 6 = 6 A = ( x 2 + 2 x − 1 ) 2 + 6 = 6 B = 2 sin 2 4 θ tan 2 4 θ + 2 cos 2 4 θ cot 2 4 θ = 2 ( cos 2 4 θ + sin 2 4 θ ) = 2 B = \dfrac{2\sin^{2} 4\theta}{\tan^{2} 4\theta} + \dfrac{2\cos^{2} 4\theta}{\cot^{2} 4\theta} = 2\left(\cos^{2} 4\theta + \sin^{2} 4\theta\right) = 2 B = tan 2 4 θ 2 sin 2 4 θ + cot 2 4 θ 2 cos 2 4 θ = 2 ( cos 2 4 θ + sin 2 4 θ ) = 2 ⇒ A B = 12 ⇒ A B 2 = 6 ⇒ AB = 12 ⇒ \dfrac{AB}{2} = 6 ⇒ A B = 12 ⇒ 2 A B = 6
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