InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Sin-cos+1/sin+cos-1 =1/sec-tan prove |
| Answer» We have to prove that,{tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex} using identity\xa0{tex}sec^2\\theta=1+tan^2\\theta{/tex}LHS = {tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} {/tex}{tex} = \\frac{{\\tan \\theta - 1 + \\sec \\theta }}{{\\tan \\theta + 1 - \\sec \\theta }}{/tex} [ dividing the numerator and denominator by {tex}\\cos{\\theta}{/tex}.]{tex} = \\frac{{(\\tan \\theta + \\sec \\theta)-1 }}{{(\\tan \\theta - \\sec \\theta )+1}}{/tex}{tex}=\\frac{\\{{(\\tan\\theta+\\sec\\theta)-1\\}}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [ Multiplying and dividing by {tex}(\\tan{\\theta}-\\sec{\\theta}){/tex}]{tex}=\\frac{{(\\tan^2\\theta-\\sec^2\\theta)-}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [{tex}\\because (a-b)(a+b)=a^2-b^2{/tex}]{tex} = \\frac{{-1-\\tan \\theta + \\sec \\theta }}{{(\\tan \\theta - \\sec \\theta+1)(\\tan{\\theta}-\\sec{\\theta}) }}{/tex}[{tex}\\because \\tan^2\\theta-\\sec^2\\theta=-1{/tex}]{tex}=\\frac{-(\\tan\\theta-\\sec\\theta+1)}{(\\tan\\theta-\\sec\\theta+1)(\\tan\\theta-\\sec\\theta)}{/tex}{tex}=\\frac{-1}{\\tan{\\theta}-\\sec{\\theta}}{/tex}{tex} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex}=RHSHence Proved. | |