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| 1. |
(sec theta+tan theta)=1/sec theta-tan theta ) |
| Answer» L.H.S.\xa0{tex}\\sec \\theta + \\tan \\theta {/tex}Rationalizing with\xa0{tex}\\sec \\theta - \\tan \\theta {/tex}{tex}\\sec \\theta + \\tan \\theta \\times {{\\sec \\theta - \\tan \\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}=\xa0{tex}{{{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}=\xa0{tex}{1 \\over {\\sec \\theta - \\tan \\theta }}{/tex} [Since\xa0{tex}{\\sec ^2}\\theta - {\\tan ^2}\\theta = 1{/tex}] | |