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| 1. |
1+ cos theta+sin theta/1+cos theta- sin theta =1+ sin theta/cos theta |
| Answer» LHS\xa0{tex} \\frac{{1 + \\cos \\theta + \\sin \\theta }}{{1 + \\cos \\theta - \\sin \\theta }}{/tex}Dividing numerator and denominator by cos{tex} \\theta {/tex}{tex}= \\frac{{\\frac{1}{{\\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\cos \\theta }} + \\frac{{\\sin \\theta }}{{\\cos \\theta }}}}{{\\frac{1}{{\\cos \\theta }} + \\frac{{\\cos \\theta }}{{\\cos \\theta }} - \\frac{{\\sin \\theta }}{{\\cos \\theta }}}}{/tex}{tex}= \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 - \\tan \\theta }}{/tex}Multiplying and dividing by\xa0{tex} \\sec \\theta + 1 + \\tan \\theta {/tex}{tex}= \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 - \\tan \\theta }} \\times \\frac{{\\sec \\theta + 1 + \\tan \\theta }}{{\\sec \\theta + 1 + \\tan \\theta }}{/tex}{tex}= \\frac{{{{(\\sec \\theta + 1 + \\tan \\theta )}^2}}}{{{{(\\sec \\theta + 1)}^2} - {{\\tan }^2}\\theta }}{/tex}{tex}= \\frac{{{{\\sec }^2}\\theta + 1 + {{\\tan }^2}\\theta + 2\\sec \\theta + 2\\tan \\theta + 2\\sec \\theta \\tan \\theta }}{{1 + {{\\sec }^2}\\theta + 2\\sec \\theta - {{\\tan }^2}\\theta }}{/tex}Now,\xa0{tex} 1 + {\\tan ^2}\\theta = {\\sec ^2}\\theta {/tex}{tex}= \\frac{{2{{\\sec }^2}\\theta + 2\\sec \\theta + 2\\tan \\theta + 2\\sec \\theta \\tan \\theta }}{{2 + 2\\sec \\theta }}{/tex}{tex} = \\frac{{2\\left[ {\\sec \\theta (\\sec \\theta + 1) + \\tan \\theta (1 + \\sec \\theta } \\right]}}{{2(1 + \\sec \\theta )}}{/tex}{tex} = \\frac{{(\\sec + \\tan \\theta )(\\sec \\theta + 1)}}{{(1 + \\sec \\theta )}}{/tex}{tex} = \\sec \\theta + \\tan \\theta = \\frac{1}{{\\cos \\theta }} + \\frac{{\\sin \\theta }}{{\\cos \\theta }}{/tex}{tex}= \\frac{{1 + \\sin \\theta }}{{\\cos \\theta }} = RHS{/tex} | |