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If cosθ = 0.6, show that (5sinθ − 3tanθ) = 0. |
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Answer» Given that cos\(\theta\) = 0.6. Therefore sin\(\theta\) = \(\sqrt{1 − cos^2\theta}\) = \(\sqrt{1 − (0.6)^2}\) = \(\sqrt{1 − 0.36}\) = \(\sqrt{0.64 }\) = 0.8. And tan\(\theta\) = \(\frac{sin \theta}{cos \theta}\) = \(\frac{0.8}{0.6}\) =\(\frac{8}{6}\) = \(\frac{4}{3}\) . Now, 5 sin\(\theta\) – 3tan\(\theta\) = 5 × 0.8 − 3 × \(\frac{4}{3}\) = 4 – 4 = 0. Hence Proved |
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