Question

Prove the following statement.
\[\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =\dfrac{1-{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }{2+{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }\]

Answer 1

Hint: To prove this question, we should know that sec α can be written as $\dfrac{1}{\cos \alpha }$ and cosec α can be written as $\dfrac{1}{\sin \alpha }$. Also, we should know a few algebraic identities like, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. By using these identities we can prove the statement.
Complete step-by-step answer:
In this question, we have been asked to prove \[\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =\dfrac{1-{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }{2+{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }\]. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha $
Now, we know that $\sec \alpha =\dfrac{1}{\cos \alpha }$ and $\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }$. So, we will get the LHS as,
$LHS=\left( \dfrac{1}{\dfrac{1}{{{\cos }^{2}}\alpha }-{{\cos }^{2}}\alpha }+\dfrac{1}{\dfrac{1}{{{\sin }^{2}}\alpha }-{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha $
Now, we will take the LCM of each term of the LHS. So, we will get,
$\begin{align}
  & LHS=\left( \dfrac{1}{\dfrac{1-{{\cos }^{4}}\alpha }{{{\cos }^{2}}\alpha }}+\dfrac{1}{\dfrac{1-{{\sin }^{4}}\alpha }{{{\sin }^{2}}\alpha }} \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \\
 & LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{1-{{\cos }^{4}}\alpha }+\dfrac{{{\sin }^{2}}\alpha }{1-{{\sin }^{4}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}} \\
\end{align}$
We can write ${{\cos }^{4}}\alpha $ as ${{\left( {{\cos }^{2}}\alpha \right)}^{2}}$ and ${{\sin }^{4}}\alpha $ as ${{\left( {{\sin }^{2}}\alpha \right)}^{2}}$. So, we will get, $LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{1-{{\left( {{\cos }^{2}}\alpha \right)}^{2}}}+\dfrac{{{\sin }^{2}}\alpha }{1-{{\left( {{\sin }^{2}}\alpha \right)}^{2}}} \right){{\sin }^{2}}\alpha {{\cos }^{2}}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. So, on applying this in the above expression we will get,
$LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{\left( 1-{{\cos }^{2}}\alpha \right)\left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{2}}\alpha }{\left( 1-{{\sin }^{2}}\alpha \right)\left( 1+{{\sin }^{2}}\alpha \right)} \right){{\sin }^{2}}\alpha {{\cos }^{2}}$
Now, we also know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$. So, we can write \[1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha \] and $1-{{\sin }^{2}}\alpha ={{\cos }^{2}}\alpha $. Therefore, we will get the LHS as,
$LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)} \right){{\sin }^{2}}\alpha {{\cos }^{2}}$
Now, we will open the brackets to simply it,
$LHS=\left( \dfrac{{{\cos }^{4}}\alpha {{\sin }^{2}}\alpha }{{{\sin }^{2}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha }{{{\cos }^{2}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)} \right)$
We know that common terms get cancelled out, so we get,
$LHS=\dfrac{{{\cos }^{4}}\alpha }{\left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{4}}\alpha }{\left( 1+{{\sin }^{2}}\alpha \right)}$
Now, we will take the LCM of both the terms. So, we will get,
\[\begin{align}
  & LHS=\dfrac{{{\cos }^{4}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)+{{\sin }^{4}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}{\left( 1+{{\cos }^{2}}\alpha \right)\left( 1+{{\sin }^{2}}\alpha \right)} \\
 & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\cos }^{4}}\alpha {{\sin }^{2}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha }{1+{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha } \\
\end{align}\]
Now, we can see that \[{{\cos }^{4}}\alpha {{\sin }^{2}}\alpha +{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha \] can be written as \[{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)\]. So, we will get,
\[LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}{1+{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }\]
We also know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ so, we can write the LHS as,
\[\begin{align}
  & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( 1 \right)}{1+1+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\
 & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\
\end{align}\]
Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we can say that ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$. Therefore, we can write \[{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha \] as \[{{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-2{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \]. So, we will get,
\[\begin{align}
  & LHS=\dfrac{{{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\
 & LHS=\dfrac{{{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\
\end{align}\]
Now, we will again put ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$. So, we will get,
\[\begin{align}
  & LHS=\dfrac{1-{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\
 & LHS=RHS \\
\end{align}\]
Hence proved.

Note: In this question, there are high possibilities that a student may make calculation mistakes. Also, they could make a mistake while applying the algebraic and trigonometric formulas. So, the students have to be careful while solving the question.