InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
(CosecA-sinA) (secA-cosA) (tanA+cotA) =1 |
| Answer» To prove:(cosecA - sinA) (secA - cosA) (tanA + cotA) = 1LHS\xa0{tex}= (\\cos ecA - \\sin A)(\\sec A - \\cos A)(\\tan A + \\cot A){/tex}{tex} = \\left( {\\frac{1}{{\\sin A}} - \\sin A} \\right)\\left( {\\frac{1}{{\\cos A}} - \\cos A} \\right)\\left( {\\frac{{\\sin A}}{{\\cos A}} + \\frac{{\\cos A}}{{\\sin A}}} \\right){/tex}\xa0{tex}\\left[ \\begin{gathered} \\because \\cos ecA = \\frac{1}{{\\sin A}},\\sec A = \\frac{1}{{\\cos A}}, \\hfill \\\\ \\tan A = \\frac{{\\sin A}}{{\\cos A}},\\cot A = \\frac{{\\cos A}}{{\\sin A}} \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\left( {\\frac{{1 - {{\\sin }^2}A}}{{\\sin A}}} \\right)\\left( {\\frac{{1 - {{\\cos }^2}A}}{{\\cos A}}} \\right)\\left( {\\frac{{{{\\sin }^2}A + {{\\cos }^2}A}}{{\\cos A\\sin A}}} \\right){/tex}{tex} = \\frac{{{{\\cos }^2}A}}{{\\sin A}} \\times \\frac{{{{\\sin }^2}A}}{{\\cos A}} \\times \\frac{1}{{\\cos A\\sin A}}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\because 1 - {\\sin ^2}A = {\\cos ^2}A,1 - {\\cos ^2}A = {\\sin ^2}A, \\hfill \\\\ {\\sin ^2}A + {\\cos ^2}A = 1 \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{{\\cos A \\times \\cos A}}{{\\sin A}} \\times \\frac{{\\sin A \\times \\sin A}}{{\\cos A}} \\times \\frac{1}{{\\cos A\\sin A}}{/tex}= 1= RHS | |