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# If we have an expression as $x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }$, then $\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta }$ is equal to:A. $1+x$ B. $1-x$ C. $x$ D. $\dfrac{1}{x}$

Last updated date: 13th Jul 2024
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According to our question it is asked of us to find the correct option for the value of $f\left( x \right)$. As we know that if any question is asking for the values of trigonometric functions, then we directly put the value of that function there and get our required answer for that. But if it is given to find the angles of them, then we find the original function (trigonometric function) for that value and then we get the angle of that by comparing it with each other. And if it is asked directly without any angle, then we always use the general formulas of trigonometry for solving this. Since,
$x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }$
\begin{align} &\Rightarrow x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }\times \dfrac{1-\cos \theta +\sin \theta }{1-\cos \theta +\sin \theta } \\ & \because \left[ \left( 1+\sin \theta \right)+\cos \theta \right]\left[ \left( 1+\sin \theta \right)-\cos \theta \right] \\ &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{{{\left( 1+\sin \theta \right)}^{2}}-{{\cos }^{2}}\theta } \\ &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{{{\left( 1+\sin \theta \right)}^{2}}-\left( 1-{{\sin }^{2}}\theta \right)} \\ &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{\left( 1+\sin \theta \right)\left[ 1+\sin \theta -\left( 1-\sin \theta \right) \right]} \\ &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{\left( 1+\sin \theta \right).2\sin \theta } \\ &\Rightarrow x=\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta } \\ \end{align}