Question

Prove the following statements.
$\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}$

Answer 1

Hint: In order to prove the given relation, we should have some knowledge regarding the trigonometric ratio transformations like $\operatorname{cosecA}=\dfrac{1}{sinA}$ and $\cot A=\dfrac{\cos A}{\sin A}$. Also, we should know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. By using these properties, we will get the desired result.
Complete step-by-step answer:
In this question, we have been asked to prove that $\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}$. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\dfrac{1}{cosecA-cotA}-\dfrac{1}{sinA}$
Now, we know that $\operatorname{cosecA}=\dfrac{1}{sinA}$ and $\cot A=\dfrac{\cos A}{\sin A}$. So, we can write the LHS as,
$LHS=\dfrac{1}{\dfrac{1}{\sin A}-\dfrac{\cos A}{\sin A}}-\dfrac{1}{\sin A}$
Now, we will take LCM of the terms $\dfrac{1}{\sin A}$ and $\dfrac{\cos A}{\sin A}$. So, we will write the LHS as,
$LHS=\dfrac{1}{\dfrac{1-\cos A}{\sin A}}-\dfrac{1}{\sin A}$
We can further write it as,
$LHS=\dfrac{\sin A}{1-\cos A}-\dfrac{1}{\sin A}$
Now, we will multiply the numerator and the denominator of $\dfrac{\sin A}{1-\cos A}$ by $\left( 1+\cos A \right)$ to rationalize $\left( 1-\cos A \right)$. So, we get,
\[LHS=\dfrac{\sin A\left( 1+\cos A \right)}{\left( 1-\cos A \right)\left( 1+\cos A \right)}-\dfrac{1}{\sin A}\]
And we know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. So, we can write $\left( 1-\cos A \right)\left( 1+\cos A \right)=1-{{\cos }^{2}}A$. Therefore, we will get,
\[LHS=\dfrac{\sin A\left( 1+\cos A \right)}{1-{{\cos }^{2}}A}-\dfrac{1}{\sin A}\]
Now, we know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. So, we can write $1-{{\cos }^{2}}A={{\sin }^{2}}A$. Therefore we will get the LHS as follows.
\[LHS=\dfrac{\sin A\left( 1+\cos A \right)}{{{\sin }^{2}}A}-\dfrac{1}{\sin A}\]
We know that the common terms in the numerator and denominator will get cancelled out. So, we get,
\[LHS=\dfrac{\left( 1+\cos A \right)}{\sin A}-\dfrac{1}{\sin A}\]
Now, we will take the LCM , so we get,
\[LHS=\dfrac{1+\cos A-1}{\sin A}\]
Now we will take the negative sign common from (cos A – 1). So, we get,
\[\begin{align}
  & LHS=\dfrac{1-\left( -\cos A+1 \right)}{\sin A} \\
 & LHS=\dfrac{1-\left( 1-\cos A \right)}{\sin A} \\
\end{align}\]
Now, we will separate each term of numerator with the denominator. So, we will get,
\[LHS=\dfrac{1}{\sin A}-\dfrac{\left( 1-\cos A \right)}{\sin A}\]
   Now, we will multiply the numerator and the denominator of \[\dfrac{\left( 1-\cos A \right)}{\sin A}\] by (1 + cos A). So, we get,
 \[LHS=\dfrac{1}{\sin A}-\dfrac{\left( 1-\cos A \right)\left( 1+\cos A \right)}{\sin A\left( 1+\cos A \right)}\]
Bu using the property $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ we can write $\left( 1-\cos A \right)\left( 1+\cos A \right)=1-{{\cos }^{2}}A$. Therefore, we will get,
\[LHS=\dfrac{1}{\sin A}-\dfrac{1-{{\cos }^{2}}A}{\sin A\left( 1+\cos A \right)}\]
We know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. So, we can write $1-{{\cos }^{2}}A={{\sin }^{2}}A$. Therefore we will get,
\[LHS=\dfrac{1}{\sin A}-\dfrac{{{\sin }^{2}}A}{\sin A\left( 1+\cos A \right)}\]
Here, we will cancel sin A from the numerator and the denominator as it is a common term. So, we get,
\[LHS=\dfrac{1}{\sin A}-\dfrac{\sin A}{\left( 1+\cos A \right)}\]
We can further write it as,
\[\begin{align}
  & LHS=\dfrac{1}{\sin A}-\dfrac{1}{\dfrac{1+\cos A}{\sin A}} \\
 & LHS=\dfrac{1}{\sin A}-\dfrac{1}{\dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}} \\
\end{align}\]
We know that $\dfrac{1}{sinA}=\operatorname{cosecA}$ and $\dfrac{\cos A}{\sin A}=\cot A$. So, we can write the LHS as,
$LHS=\dfrac{1}{sinA}-\dfrac{1}{cosecA+cotA}$
$LHS=RHS$
Hence proved.

Note: In this question, there are high possibilities that the students may get stuck at the step where we take the negative sign out. So, in such cases, they can simply start from the RHS of the given relation and then, by solving the RHS, they will be able to get LHS = RHS.