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CosA-sinA+1/cosA+sinA-1=cosecA+cotA |
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Answer» Thank you so much. {tex}{{\\cos {\\rm{A}} - \\sin {\\rm{A}} + 1} \\over {\\cos {\\rm{A}} + \\sin {\\rm{A}} - 1}}{/tex}Dividing all terms by sin A, we get{tex}{{\\cot {\\rm{A}} - 1 + \\cos ec{\\rm{A}}} \\over {\\cot {\\rm{A}} + 1 - \\cos ec{\\rm{A}}}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} - 1} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} - \\left( {\\cos e{c^2}{\\rm{A}} - {{\\cot }^2}{\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}\\left( {1 - \\cos ec{\\rm{A}} + \\cot {\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}}{/tex}Hence proved cosA-sinA+1/cosA+sinA-1= 1/sinA+cosA/sinAcosA-sinA+1/cosA+sinA-1=sinA+cosAsinA/sin (square)AcosA-sinA+1/cosA+sinA-1=sina(1+cosA)/sin(square)AcosA-sinA+1/cosA+sinA-1=1+cosA/sinAsinAcosA-sin(square)A+sinA=cosA+sinA-1+cos(square)A+sinAcosA-cosA(by cross multiplication)sincosA-sin(square)A+sinA=cosA+sinA-1+1-sin(square)A+sinAcosA-cosA0=0Hence proved Yes |
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