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Without using trigonometric tables, show that \(\frac{cos\,70°}{sin\,20°}\)+ cos 49° sin 41° = 2 |
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Answer» \(\frac{cos\,70°}{sin\,20°}\)+ cos 49°cosec 41° = \(\frac{cos(90°-20°)}{sin\,20°}+\)cos(90° - 41°)cosec 41° = \(\frac{sin\,20°}{sin\,20°}+\) sin 41° x \(\frac1{sin\,41°}\) = 1 + 1 = 2. |
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