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| 1. |
(1+tan×tan A)+(1+1/tan×tan A)=1/sin×sinA-sin×sin×sin×sinA |
| Answer» LHS{tex} = \\left( {1 + {{\\tan }^2}A} \\right) + \\left( {1 + \\frac{1}{{{{\\tan }^2}A}}} \\right){/tex}{tex} = \\left( {1 + {{\\tan }^2}A} \\right) + \\frac{{({{\\tan }^2}A + 1)}}{{{{\\tan }^2}A}}{/tex}{tex} = {\\sec ^2}A + \\frac{{{{\\sec }^2}A}}{{{{\\tan }^2}A}}{/tex}\xa0{tex}\\left[ {\\because 1 + {{\\tan }^2}A = {{\\sec }^2}A} \\right]{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{{\\frac{1}{{{{\\cos }^2}A}}}}{{\\frac{{{{\\sin }^2}A}}{{{{\\cos }^2}A}}}}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\because {\\sec ^2}A = \\frac{1}{{{{\\cos }^2}A}} \\hfill \\\\ {\\tan ^2}A = \\frac{{{{\\sin }^2}A}}{{{{\\cos }^2}A}} \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{1}{{{{\\cos }^2}A}} \\times \\frac{{{{\\cos }^2}A}}{{{{\\sin }^2}A}}{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{1}{{{{\\sin }^2}A}}{/tex}{tex} = \\frac{{{{\\sin }^2}A + {{\\cos }^2}A}}{{{{\\cos }^2}A{{\\sin }^2}A}}{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A{{\\sin }^2}A}}{/tex}\xa0{tex}\\left[ {\\because {{\\sin }^2}A + {{\\cos }^2}A = 1} \\right]{/tex}{tex} = \\frac{1}{{(1 - {{\\sin }^2}A){{\\sin }^2}A}}{/tex}\xa0{tex}\\left[ {\\because {{\\cos }^2}A = 1 - {{\\sin }^2}A} \\right]{/tex}{tex} = \\frac{1}{{{{\\sin }^2}A - {{\\sin }^4}A}}{/tex}= RHSHence proved. | |