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If Sec theta +tan theta = p, find cosec theta

Answer» Given,{tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also, we know that,\xa0{tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1 [{tex}\\because a^2-b^2=(a+b)(a-b){/tex}]{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex})p = 1 [using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(i)+(ii), we get,{tex}sec\\theta + tan\\theta+ sec\\theta - tan\\theta = p+ \\frac{1}{p}{/tex}{tex}\\Rightarrow 2sec\\theta = \\frac{p^2+1}{p}{/tex}{tex}\\Rightarrow sec\\theta = \\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow \\frac{1}{cos\\theta} =\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow cos\\theta =\\frac{2p}{p^2+1}{/tex}------(iii)Now, we know that,{tex}sin\\theta = \\sqrt( 1- cos^2\\theta) {/tex}put the value of\xa0{tex}cos\\theta{/tex}\xa0from eq. (iii), we get,{tex}sin\\theta = \\sqrt(1-(\\frac{2p}{p^2+1})^2){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(1-\\frac{4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1+2p^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1-2p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2-1)^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\frac{p^2-1}{p^2+1}{/tex}{tex}cosec\\theta = \\frac{p^2+1}{p^2-1} [\\because cosec\\theta =\\frac{1}{sin\\theta}]{/tex}hence, {tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex}


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