InterviewSolution
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InterviewSolution
| 1. |
Express the cosA in terms of cotA |
| Answer» tan A =\xa0{tex}=\\frac{1}{\\cot A}{/tex}{tex}\\Rightarrow{/tex}\xa0tan2\xa0A =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec2\xa0A - 1 =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec2\xa0A =\xa0{tex}\\frac{1}{\\cot ^{2} A}{/tex}+1{tex}\\Rightarrow{/tex}\xa0sec2\xa0A =\xa0{tex}\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}{/tex}{tex}\\Rightarrow{/tex}\xa0sec A =\xa0{tex}\\sqrt{\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{\\cos A}=\\sqrt{\\frac{1+\\cot ^{2} A}{\\cot ^{2} A}}{/tex}{tex}\\Rightarrow{/tex}\xa0cos A =\xa0{tex}\\sqrt{\\frac{\\cot ^{2} A}{1+\\cot ^{2} A}}{/tex}\xa0{tex}=\\frac{\\cot A}{\\sqrt{1+\\cot ^{2} A}}{/tex}{tex}\\therefore{/tex}\xa0cos A =\xa0{tex}\\frac{\\cot A}{\\sqrt{1+\\cot ^{2} A}}{/tex} | |