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| 1. |
(√3-1)(3-cot30°)=tan^30°-2sin^260° |
| Answer» We have to prove that:\xa0{tex}( \\sqrt { 3 } + 1 ){/tex}\xa0(3 – cot 30°) = tan3\xa060° – 2 tan 60°Here, LHS =\xa0{tex}( \\sqrt { 3 } + 1 ){/tex}\xa0(3 – cot 30°){tex}= ( \\sqrt { 3 } + 1 ) ( 3 - \\sqrt { 3 } ){/tex}{tex}= \\sqrt { 3 } ( 3 - \\sqrt { 3 } ) + 1 ( 3 - \\sqrt { 3 } ){/tex}{tex}= 3 \\sqrt { 3 } - 3 + 3 - \\sqrt { 3 }{/tex}{tex}= 2 \\sqrt { 3 }{/tex}RHS = tan3\xa060° – 2 sin 60°{tex}= ( \\sqrt { 3 } ) ^ { 3 } - 2 \\times \\frac { \\sqrt { 3 } } { 2 }{/tex}{tex}= 3 \\sqrt { 3 } - \\sqrt { 3 }{/tex}{tex}= 2 \\sqrt { 3 }{/tex}⇒ LHS = RHSHence, proved. | |