If \({a^2} = \frac{{1 + 2\sin \theta \cos \theta }}{{1 - 2\sin \theta \cos \theta }}\), then what is the value of \(\frac{{\frac{1}{a} + 1}}{{\frac{1}{a} - 1}}\)?
If \({a^2} = \frac{{1 + 2\sin \theta \cos \theta }}{{1 - 2\sin \theta \cos \theta }}\), then what is the value of \(\frac{{\frac{1}{a} + 1}}{{\frac{1}{a} - 1}}\)?
1). secθ
2). 1
3). -1
4). -tanθ
1 answers
Given,
$(\begin{array}{l} {a^2} = \frac{{1 + 2\sin \theta \cos \theta }}{{1 - 2\sin \theta \cos \theta }}\\ \Rightarrow {a^2} = \frac{{si{n^2}\theta + co{s^2}\theta + 2\;sin\theta cos\theta }}{{si{n^2}\theta + co{s^2}\theta - 2sin\theta cos\theta }}\\ \Rightarrow {a^2} = \frac{{{{\left( {sin\theta + cos\theta } \right)}^2}}}{{{{\left( {sin\theta - cos\theta } \right)}^2}}} \end{array})$
$(\Rightarrow a = \frac{{sin\theta + cos\theta }}{{sin\theta - cos\theta }})$ ......(1)
Now consider,
$(\begin{array}{l} \frac{{\frac{1}{a} + 1}}{{\frac{1}{a} - 1}}\\ \Rightarrow \frac{{1 + a}}{{1 - a}} \end{array})$
From (1), we get
$(\begin{array}{l} \Rightarrow \frac{{1 + \frac{{sin\theta + cos\theta }}{{sin\theta - cos\theta }}}}{{1 - \frac{{sin\theta + cos\theta }}{{sin\theta - cos\theta }}}}\\ \Rightarrow \frac{{sin\theta - cos\theta + sin\theta + cos\theta }}{{sin\theta - cos\theta - sin\theta - cos\theta }}\\ \Rightarrow \frac{{2sin\theta }}{{ - 2cos\theta }} \end{array})$
$(\Rightarrow - tan\theta \left[ {? tan\theta = \frac{{sin\theta }}{{cos\theta }}} \right])$
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