If Sec theta +tan theta = …

Given,

{tex}sec\ \theta+ tan\ \theta  = p{/tex} ...(i)
Also, we know that,

 {tex}sec^2 \theta  - tan^2 \theta  = 1{/tex}
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) (sec {tex}\theta{/tex} + tan {tex}\theta{/tex}) = 1 [{tex}\because a^2-b^2=(a+b)(a-b){/tex}]
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex})p = 1  [using equation (i)]
{tex}\Rightarrow{/tex} sec {tex}\theta{/tex} - tan {tex}\theta{/tex} {tex}=\frac{1}{p}{/tex} ...(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 {tex}cos\theta{/tex} from 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\ \theta{/tex} {tex}=\frac{1+p^{2}}{1-p^{2}}{/tex}