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wolverine [178]
2 years ago
7

%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">
pls i want the answwr as fast u can
ASAP !


MisterBrainly Answer Please :((((
0r any top user
who will answer correctly he will be a good user
plz answer​
Mathematics
1 answer:
inysia [295]2 years ago
5 0

\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}

<h2>Hence,</h2>

\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}

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