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Without using trigonometric tables, prove that(i) \(\frac{sin70°}{cos20°}+\frac{cosec20°}{sec70°}-2cos70°cosec20°= 0\)(ii) \(\frac{cos80°}{sin10°}+cos59°cosec31° = 2\)(iii) \(\frac{sin18°}{sin72°}+\sqrt3(tan10°tan30°tan40°tan50°tan80°) = 2°\) |
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Answer» (i) LHS = \(\frac{sin70°}{cos20°}+\frac{cosec20°}{sec70°}-2cos70°cosec20°\) = \(\frac{sin70°}{sin(90° - 20°)}+\frac{sec(90°-20°)}{sec70°}- 2cos70°sec(90°-20°)\) = \(\frac{sin70°}{sin70°}+\frac{sec70°}{sec70°}-2cos70°sec70°\) = 1 + 1 - 2 x cos70° x\(\frac{1}{cos70°}\) = 2 - 2 = 0 = RHs (ii) LHS = \(\frac{cos80°}{sin10°}+cos59°cosec31°\) = \(\frac{cos80°}{sin(90°-10°)}+sin(90°-59°)cosec31°\) = \(\frac{cos80°}{cos80°}+sin31°cosec31°\) = \(\frac{cos80°}{sin10°}+cos59°cosec31°\) = 1 + sin31° x \(\frac{1}{sin31°}\)= 1 + 1 = 2 = RHS (iii) LHS = \(\frac{sin18°}{sin72°}+\sqrt3(tan10°tan30°tan40°tan50°tan80°) \) = \(\frac{sin18°}{sin90°-72°}+\sqrt3[cot(90°-10°)\times\frac{1}{\sqrt3}\times{cot(90°-40°)}\times{tan50°}\times{tan80°}]\) = \(\frac{sin18°}{sin18°}+ \sqrt3(\frac{cos80°\times{cot50°}\times{tan50°}\times{tan80°})}{\sqrt3}\) = 1 + 1 = 2 = RHS |
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