If sinA=√3/2 find value of 2cotsquare-1
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According to the question, sin A\xa0{tex}= \\frac { \\sqrt { 3 } } { 2 }{/tex}{tex}sin A= sin 60°{/tex}{tex}\\Rightarrow{/tex}\xa0A = 60°Now 2 cot2\xa0A – 1\xa0= 2cot2\xa060° – 1{tex}= 2 \\times \\left( \\frac { 1 } { \\sqrt { 3 } } \\right) ^ { 2 }{/tex}\xa0- 1{tex}= – \\frac { 1 } { 3 }{/tex}