InterviewSolution
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InterviewSolution
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What is the value of \(\frac{{\left( {cos10^\circ + sin20^\circ } \right)}}{{\left( {cos20^\circ - sin10^\circ } \right)}}?\)1). \(\frac{1}{{\sqrt 3 }}\)2). \(- \frac{1}{{\sqrt 3 }}\)3). \(\sqrt 3\)4). \(- \sqrt 3\) |
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Answer» Concept: $(\begin{ARRAY}{L} \COS \left( {90 - \theta } \right) = \sin \theta \\ \sin \left( {90 - \theta } \right) = \cos \theta \END{array})$ $(\sin A + \sin B = 2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right))$ $(\sin A - \sin B = 2\sin \left( {\frac{{A - B}}{2}} \right)\cos \left( {\frac{{A + B}}{2}} \right))$ Calculation: $(\frac{{\cos 10^\circ + \sin 20^\circ }}{{\cos 20^\circ - \sin 10^\circ }} = \frac{{\cos \left( {90^\circ - 80^\circ } \right) + \sin 20^\circ }}{{\cos \left( {90^\circ - 70^\circ } \right) - \sin 10^\circ }})$ $(= \frac{{\sin 80^\circ + \sin 20^\circ }}{{\sin 70^\circ - \sin 10^\circ }} = \frac{{2\sin 50^\circ \cos 30^\circ }}{{2\cos 40^\circ \sin 30^\circ }})$ $(= \frac{{\sin \left( {90^\circ - 40^\circ } \right)\cot 30^\circ }}{{\cos 40^\circ }})$ $(= \frac{{\cos 40^\circ \cot 30^\circ }}{{\cos 40^\circ }} = \cot 30^\circ = \sqrt 3 )$ |
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