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| 1. |
Sinthita - costhita + 1 ÷ sinthita + costhita - 1 = 1 ÷ secthita - tanthita |
| Answer» L.H.S. = {tex}{{\\sin \\theta - \\cos \\theta + 1} \\over {\\sin \\theta + \\cos \\theta - 1}}{/tex}Dividing all terms by {tex}\\cos \\theta {/tex}= {tex}{{\\tan \\theta - 1 + \\sec \\theta } \\over {\\tan \\theta + 1 - \\sec \\theta }}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - 1} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - \\left( {{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - \\left( {\\sec \\theta + \\tan \\theta } \\right)\\left( {\\sec \\theta - \\tan \\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\left( {\\tan \\theta + \\sec \\theta } \\right)\\left( {1 - \\sec \\theta + \\tan \\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}\\sec \\theta + \\tan \\theta {/tex}= {tex}\\sec \\theta + \\tan \\theta \\times {{\\sec \\theta - \\tan \\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}= {tex}{{{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}= {tex}{1 \\over {\\sec \\theta - \\tan \\theta }}{/tex}= R.H.S. | |