Question

%20%5Ctan%20%5Ctheta%20%2B%20%5Csec%20%5Ctheta%20-%201%20%7D%20%7B%20%5Ctan%20%5Ctheta%20-%20%5Csec%20%5Ctheta%20%2B%201%20%7D%20%3D%20%5Cfrac%20%7B%201%20%2B%20%5Csin%20%5Ctheta%20%7D%20%7B%20%5Ccos%20%5Ctheta%20%7D" id="TexFormula1" title="prove that\ \textless \ br /\ \textgreater \ \frac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { 1 + \sin \theta } { \cos \theta }" alt="prove that\ \textless \ br /\ \textgreater \ \frac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { 1 + \sin \theta } { \cos \theta }" align="absmiddle" class="latex-formula">
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Answer 1

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

Hence,

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$