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| 1. |
Prove that 1+tan^2x/1+cot^2x=(1-tanx/1-cotx)^2 |
| Answer» To prove :\xa0{tex} \\frac{{1 + {{\\tan }^2}\\theta }}{{1 + {{\\cot }^2}\\theta }} = {\\left( {\\frac{{1 - \\tan \\theta }}{{1 - \\cot \\theta }}} \\right)^2}{/tex}Consider : {tex} \\frac{{1 + {{\\tan }^2}\\theta }}{{1 + {{\\cot }^2}\\theta }} = \\frac{{1 + \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}}}{{1 + \\frac{{{{\\cos }^2}\\theta }}{{{{\\sin }^2}\\theta }}}}=\\frac{{\\frac{{{{\\cos }^2}\\theta + {{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}}}{{\\frac{{{{\\sin }^2}\\theta + {{\\cos }^2}\\theta }}{{{{\\sin }^2}\\theta }}}}{/tex}{tex}= \\frac{{\\frac{1}{{{{\\cos }^2}\\theta }}}}{{\\frac{1}{{{{\\sin }^2}\\theta }}}} = \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }}{/tex}\xa0{tex} \\left[ {\\because {{\\sin }^2}\\theta + {{\\cos }^2}\\theta = 1} \\right]{/tex}{tex}= {\\tan ^2}\\theta {/tex}Consider\xa0{tex} {\\left( {\\frac{{1 - \\tan \\theta }}{{1 - \\cot \\theta }}} \\right)^2} = \\frac{{1 + {{\\tan }^2}\\theta - 2\\tan \\theta }}{{1 + {{\\cot }^2}\\theta - 2\\cot \\theta }}{/tex}{tex}= \\frac{{{{\\sec }^2}\\theta - 2\\tan \\theta }}{{\\cos e{c^2}\\theta - 2\\cot \\theta }}{/tex}\xa0{tex} \\left[ {\\because 1 + {{\\tan }^2}\\theta = {{\\sec }^2}\\theta } \\right]{/tex}{tex}= \\frac{{\\frac{1}{{{{\\cos }^2}\\theta }} - \\frac{{2\\sin \\theta }}{{\\cos \\theta }}}}{{\\frac{1}{{{{\\sin }^2}\\theta }} - \\frac{{2\\cos \\theta }}{{\\sin \\theta }}}} = \\frac{{\\frac{{1 - 2\\sin \\theta \\cos \\theta }}{{{{\\cos }^2}\\theta }}}}{{\\frac{{1 - 2\\sin \\theta \\cdot \\cos \\theta }}{{{{\\sin }^2}\\theta }}}}{/tex}{tex} = \\frac{{{{\\sin }^2}\\theta }}{{{{\\cos }^2}\\theta }} = {\\tan ^2}\\theta {/tex} | |