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| 1. |
1/sec a -tan a - 1/cos a = 1/cos a - 1/seca + tan a |
| Answer» \xa0We have,\xa0{tex} \\frac{1}{{\\sec A + \\tan A}} - \\frac{1}{{\\cos A}} = \\frac{1}{{\\cos A}} - \\frac{1}{{\\sec A - \\tan A}}{/tex}{tex}\\Rightarrow \\frac{1}{{\\sec A + \\tan A}} + \\frac{1}{{\\sec A - \\tan A}} = \\frac{1}{{\\cos A}} + \\frac{1}{{\\cos A}}{/tex}LHS\xa0{tex}= \\frac{1}{{\\sec A + \\tan A}} + \\frac{1}{{\\sec A - \\tan A}}{/tex}{tex}= \\frac{{\\sec A - \\tan A + \\sec A + \\tan A}}{{(\\sec A + \\tan A)(\\sec A - \\tan A)}}{/tex}{tex}= \\frac{{2\\sec A}}{{{{\\sec }^2}A - {{\\tan }^2}A}}{/tex}\xa0{tex} \\left[ {\\because (a + b)(a - b) = ({a^2} - {b^2}} \\right]{/tex}{tex}= \\frac{{2\\sec A}}{1}{/tex}\xa0{tex} \\left[ {\\because {{\\sec }^2}A - {{\\tan }^2}A = 1} \\right]{/tex}= 2secARHS\xa0{tex}= \\frac{1}{{\\cos A}} + \\frac{1}{{\\cos A}}{/tex}{tex}= \\frac{{1 + 1}}{{\\cos A}}{/tex}{tex}= \\frac{2}{{\\cos A}}{/tex}= 2 secALHS = RHS\xa0 | |