Question

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. $\dfrac{1}{x}$

Answer 1

Hint: For solving this question you should know about finding the values of trigonometric functions. In this problem we can find the trigonometric values of the given functions by using the formulas. But if it is given to find the value of a trigonometric function, then we find the trigonometric value of that function and then we add them according to the given sequence.

Complete step-by-step solution:
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,
$x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }$
By doing rationalisation here we get,
$\begin{align}
  &\Rightarrow x=\dfrac{2\sin \theta }{1+\cos \theta +\sin \theta }\times \dfrac{1-\cos \theta +\sin \theta }{1-\cos \theta +\sin \theta } \\
 & \because \left[ \left( 1+\sin \theta \right)+\cos \theta \right]\left[ \left( 1+\sin \theta \right)-\cos \theta \right] \\
 &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{{{\left( 1+\sin \theta \right)}^{2}}-{{\cos }^{2}}\theta } \\
 &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{{{\left( 1+\sin \theta \right)}^{2}}-\left( 1-{{\sin }^{2}}\theta \right)} \\
 &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{\left( 1+\sin \theta \right)\left[ 1+\sin \theta -\left( 1-\sin \theta \right) \right]} \\
 &\Rightarrow x=\dfrac{2\sin \theta \left( 1-\cos \theta +\sin \theta \right)}{\left( 1+\sin \theta \right).2\sin \theta } \\
 &\Rightarrow x=\dfrac{1-\cos \theta +\sin \theta }{1+\sin \theta } \\
\end{align}$
So, the correct option is option C.

Note: While solving these type of problems you have to always keep in mind that if the value of inverse trigonometric function is asked, then find that and if it is asked for the angle, then find it by comparing, if that is asking for the trigonometric function value to solve any problem, then give that, but if no angles are given, then always use the general trigonometric formulas.