Question

If \[\pi <\alpha <2\pi \] then \[\dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha -{{\cos }^{2}}\alpha }}=\]
A. \[\sin \alpha \]
B. $\dfrac{-\sin \alpha }{\cos 2\alpha }$
C. \[\dfrac{1}{\sin \alpha }\]
D. 1

Answer 1

Hint: We will first start by using the fact that $\cot x=\dfrac{\cos x}{\sin x}$. Then we will take ${{\cos }^{2}}x$ a common in the square root in the denominator. Then we will use the fact that $1-{{\sin }^{2}}\alpha ={{\cot }^{2}}\alpha $ to further solve the prove, then finally we will use the fact that $\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $ to find the answer.

Complete step by step solution:
Now, we have to find the value of \[\dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha -{{\cos }^{2}}\alpha }}\].
Now, we will take ${{\cot }^{2}}\alpha $common inside the square root. So, we have,
$\Rightarrow \dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha \left( 1-\dfrac{{{\cos }^{2}}\alpha }{{{\cot }^{2}}\alpha } \right)}}$
Now, we know that ${{\cot }^{2}}\alpha =\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }$.
$\begin{align}
  & \Rightarrow \dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha \left( 1-\dfrac{{{\cos }^{2}}\alpha }{\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }} \right)}} \\
 & \Rightarrow \dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha \left( 1-{{\sin }^{2}}\alpha \right)}} \\
\end{align}$
Now, we know that $1-{{\sin }^{2}}\alpha ={{\cos }^{2}}\alpha $.
\[\begin{align}
  & \Rightarrow \dfrac{1}{\sin \alpha -\sqrt{{{\cot }^{2}}\alpha \times \left( {{\cos }^{2}}\alpha \right)}} \\
 & \Rightarrow \dfrac{1}{\sin \alpha -\cot \alpha \cos \alpha } \\
\end{align}\]
Now, again we will use the fact that $\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }$.
$\begin{align}
  & \Rightarrow \dfrac{1}{\sin \alpha -\dfrac{{{\cos }^{2}}\alpha }{\sin \alpha }} \\
 & \Rightarrow \dfrac{\sin \alpha }{{{\sin }^{2}}\alpha -{{\cos }^{2}}\alpha } \\
 & \Rightarrow \dfrac{-\sin \alpha }{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha } \\
\end{align}$
Now, we know the fact that $\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $.
$\Rightarrow \dfrac{-\sin \alpha }{\cos 2\alpha }$
Hence, the correct option is (B).

Note: It is important to note that to solve these question it is important to note that we have first used the identity ${{\cot }^{2}}\alpha =\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }$ to convert the expression inside the square root to ${{\cot }^{2}}\alpha {{\cos }^{2}}\alpha $. Also, it is important to remember the trigonometric identity that ${{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A$ to solve the problem completely.