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| 1. |
If 4 tan thita=3,evaluate 4 sin thita-cos thita+1/4 sin thita+cos thita-1 |
| Answer» We have, 4 tan\xa0{tex}\\theta{/tex}\xa0= 3{tex}\\Rightarrow{/tex}\xa0tan\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\frac{3}{4}{/tex}{tex}\\because sec\\theta=\\sqrt{1+tan^2\\theta}{/tex}\xa0=\xa0{tex}\\sqrt{1+\\frac{9}{16}}{/tex}\xa0={tex}\\sqrt{\\frac{16+9}{16}} {/tex}\xa0{tex} = \\sqrt{\\frac{25}{16}}{/tex}\xa0{tex} = \\frac{5}{4}{/tex}Now, {tex}\\frac{4 \\sin \\theta-\\cos \\theta+1}{4 \\sin \\theta+\\cos \\theta-1}{/tex}\xa0=\xa0{tex}\\frac{cos\\theta(4 \\frac{\\sin \\theta}{cos\\theta}-\\frac{\\cos \\theta}{cos\\theta}+\\frac{1}{cos\\theta})}{cos\\theta(4 \\frac{\\sin \\theta}{{cos\\theta}{}}+\\frac{\\cos \\theta}{cos\\theta}-\\frac{1}{cos\\theta})}{/tex}=\xa0{tex}\\frac{4tan\\theta-1+sec\\theta}{4tan\\theta+1-sec\\theta}{/tex}Substituting the values, we get,={tex}\\frac{4(\\frac{3}{4})-1+(\\frac{5}{4})}{4(\\frac{3}{4})+1-(\\frac{5}{4})}{/tex}\xa0= {tex}\\frac{3-1+(\\frac{5}{4})}{3+1-(\\frac{5}{4})}{/tex}\xa0=\xa0{tex}\\frac{2+(\\frac{5}{4})}{4-(\\frac{5}{4})}{/tex}\xa0=\xa0{tex}\\frac{\\frac{8+5}{4}}{\\frac{16-5}{4}}{/tex}\xa0=\xa0{tex}\\frac{\\frac{13}{4}}{\\frac{11}{4}}{/tex}\xa0=\xa0{tex}{\\frac{13}{4}}\\times{\\frac{4}{11}}{/tex}\xa0=\xa0{tex}\\frac{13}{11}{/tex} | |