InterviewSolution
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Prove that: (i) tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0.(ii) 2 sin2 \(\frac{\pi}{6}\) + cosec2 \(\frac{7\pi}{6}\) cos2 \(\frac{\pi}{3}\)= \(\frac{3}{2}\)(iii) sec(\(\frac{3\pi}{2}-\theta\)) sec(\(\theta-\frac{5\pi}{2}\)) + tan(\(\frac{5\pi}{2}+\theta\)) tan(\(\theta-\frac{5\pi}{2}\)) = - 1 |
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Answer» (i) tan(-225°) = -(tan 225°) = -(tan(180° + 45°)) = – tan 45° = – 1 cot(-405°) = -(cot 405°) = – cot(360° + 45°) [∵ For 360° + 45° no change in T-ratio.] = -cot 45° = -1 tan(-765°) = -tan 765° = -tan(2 x 360° + 45°) = -tan 45° = -1 cot 675° = cot (360°+ 315°) = cot 315° = cot(360° – 45°) = -cot 45° = -1 LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°) = (-1) (-1) – (-1) (-1) = 1 – 1 = 0 = RHS. Hence proved. (ii) 2 sin2 \(\frac{\pi}{6}\) + cosec2 \(\frac{7\pi}{6}\) cos2 \(\frac{\pi}{3}\)= \(\frac{3}{2}\) LHS = 2 sin2 \(\frac{\pi}{6}\) + cosec2 \(\frac{7\pi}{6}\) cos2 \(\frac{\pi}{3}\)= \(\frac{3}{2}\) [∵ \(\frac{7\pi}{6}\)= 210°, 210° = 180° + 30°. For 180° + 30° no change in T-ratio. 210° lies in 3rd quadrant, cosec θ is negative.] = 2(sin\(\frac{\pi}{6}\))2 + (cosec (180° + 30°))2 (cos\(\frac{\pi}{3}\))2 = 2(\(\frac{1}{2}\))2 + (-cosec 30°)2.(\(\frac{1}{2}\))2 = 2 x \(\frac{1}{4}+(-2)^2\frac{1}{4}\) = \(\frac{2}{4}+\frac{4}{4}=\frac{6}{4}\) = \(\frac{6}{4}\) = \(\frac{3}{2}\) = RHS (iii) sec(\(\frac{3\pi}{2}-\theta\)) = sec (270° – θ) = -cosec θ [∵ For 270° – θ change T-ratio. So add ‘co’ infront ‘sec’, it becomes ‘cosec’] sec(θ – \(\frac{5\pi}{2}\)) = sec(-(\(\frac{5\pi}{2}\) - θ)) = sec(\(\frac{5\pi}{2}\) – θ) [∵ sec(-θ) = θ] = sec(450° – θ) = sec (360° + (90° – θ)) = sec (90° – θ) = cosec θ [∵ For 90° – θ change in T-ratio. So add ‘co’ in front of ‘sec’ it becomes ‘cosec’] tan(\(\frac{5\pi}{2}\) + θ) = tan(450° + θ) [∵ For 90° + θ, change in T-ratio. So add ‘co’ in front of ‘tan’ it becomes ‘cot’] = tan (360° + (90° + θ)) = tan (90° + θ) = -cot θ tan(\(\theta-\frac{5\pi}{2}\)) = tan(-(\(\frac{5\pi}{2}\) - θ)) = - tan (\(\frac{5\pi}{2}\) - θ) [∵ tan(-θ) = -tan θ] = -tan(450° – θ) = -tan(360° + (90° – θ)) = -tan(90° – θ) = -cot θ LHS = sec(\(\frac{3\pi}{2}-\theta\)) sec(\(\theta-\frac{5\pi}{2}\)) + tan(\(\frac{5\pi}{2}+\theta\)) tan(\(\theta-\frac{5\pi}{2}\)) = -cosec θ (cosec θ) + (-cot θ) (-cot θ) = -cosec2 θ + cot2 θ = -(1 + cot2 θ) + cot2 θ [∵ 1 + cot2 θ = cosec2 θ] = -1 = RHS |
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