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InterviewSolution
| 1. |
Cos (90° -A) sin ( 90°-A) /tan (90°-A)=sinA |
| Answer» {tex} LHS = \\frac{{\\cos \\left( {90^\\circ - A} \\right)\\sin \\left( {90^\\circ - A} \\right)}}{{\\tan \\left( {90^\\circ - A} \\right)}}{/tex}{tex} = \\frac{{\\sin A \\times \\cos A}}{{\\cot A}}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\cos \\left( {90^\\circ - \\theta } \\right) = \\sin \\theta \\hfill \\\\ \\sin \\left( {90^\\circ - \\theta } \\right) = \\cos \\theta \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{{\\sin A \\times \\cos A}}{{\\frac{{\\cos A}}{{\\sin A}}}}{/tex}{tex}= \\sin A \\times \\cos A \\times \\frac{{\\sin A}}{{\\cos A}}{/tex}{tex}= sin^2A = RHS{/tex} | |