Question

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

Answer 1

Hint: To prove the required expression, we should know that \[\sec \theta \] can be written as \[\dfrac{1}{\cos \theta }\] and \[\operatorname{cosec}\theta \] can be written as \[\dfrac{1}{\sin \theta }\]. 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 in the relation, we can prove the desired result.

Complete step-by-step answer:

In this question, we are asked to prove that

\[\left( \dfrac{1}{{{\sec }^{2}}\theta -{{\cos }^{2}}\theta }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\theta -{{\sin }^{2}}\theta } \right){{\sin }^{2}}\theta {{\cos }^{2}}\theta =\dfrac{1-{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

To prove this expression, we will first consider the left-hand side of the expression. So, we can write it as,

\[LHS=\left( \dfrac{1}{{{\sec }^{2}}\theta -{{\cos }^{2}}\theta }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\theta -{{\sin }^{2}}\theta } \right){{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Now, we know that \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. So, we get LHS as,

\[LHS=\left( \dfrac{1}{\dfrac{1}{{{\cos }^{2}}\theta }-{{\cos }^{2}}\theta }+\dfrac{1}{\dfrac{1}{{{\sin }^{2}}\theta }-{{\sin }^{2}}\theta } \right){{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Now, we will take LCM of each term of LHS. So, we get,

\[LHS=\left[ \dfrac{1}{1-\dfrac{{{\cos }^{4}}\theta }{{{\cos }^{2}}\theta }}+\dfrac{1}{\dfrac{1-{{\sin }^{4}}\theta }{{{\sin }^{2}}\theta }} \right]{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

\[LHS=\left[ \dfrac{{{\cos }^{2}}\theta }{1-{{\cos }^{4}}\theta }+\dfrac{{{\sin }^{2}}\theta }{1-{{\sin }^{4}}\theta } \right]{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Now, we know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. So, we will get,
\[LHS=\left[ \dfrac{{{\cos }^{2}}\theta }{\left( 1-{{\cos }^{2}}\theta \right)\left( 1+{{\cos }^{2}}\theta \right)}+\dfrac{{{\sin }^{2}}\theta }{\left( 1-{{\sin }^{2}}\theta \right)\left( 1+{{\sin }^{2}}\theta \right)} \right]{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Now, we also know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, we can say that \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \] and \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \].

Therefore, we get LHS as,

\[LHS=\left[ \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta \left( 1+{{\cos }^{2}}\theta \right)}+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta \left( 1+{{\sin }^{2}}\theta \right)} \right]{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Now, we will open brackets to simplify it further. So, we get,

\[LHS=\dfrac{{{\cos }^{4}}\theta {{\sin }^{2}}\theta }{{{\sin }^{2}}\theta \left( 1+{{\cos }^{2}}\theta \right)}+\dfrac{{{\sin }^{4}}\theta {{\cos }^{2}}\theta }{{{\cos }^{2}}\theta \left( 1+{{\sin }^{2}}\theta \right)}\]

\[LHS=\dfrac{{{\cos }^{4}}\theta }{1+{{\cos }^{2}}\theta }+\dfrac{{{\sin }^{4}}\theta }{1+{{\sin }^{2}}\theta }\]

Now, we will take LCM of both the terms. So, we will get,

\[LHS=\dfrac{{{\cos }^{4}}\theta \left( 1+{{\sin }^{2}}\theta \right)+{{\sin }^{4}}\theta \left( 1+{{\cos }^{2}}\theta \right)}{\left( 1+{{\cos }^{2}}\theta \right)\left( 1+{{\sin }^{2}}\theta \right)}\]

Again, we will open all the brackets to simplify it further, so we get,

\[LHS=\dfrac{{{\cos }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{4}}\theta +{{\sin }^{4}}\theta +{{\sin }^{4}}\theta {{\cos }^{2}}\theta }{1+{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

Now, we can see that \[{{\sin }^{2}}\theta {{\cos }^{4}}\theta +{{\cos }^{2}}\theta {{\sin }^{4}}\theta \] can be written as \[{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\]. So, we get LHS as,

\[LHS=\dfrac{{{\cos }^{4}}\theta +{{\sin }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}{1+{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

We also know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Therefore, we will get LHS as,

\[LHS=\dfrac{{{\cos }^{4}}\theta +{{\sin }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( 1 \right)}{1+1+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

\[LHS=\dfrac{{{\cos }^{4}}\theta +{{\sin }^{4}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

Now, we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] which can also be written as \[{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\]. Therefore, we can write \[{{\cos }^{4}}\theta +{{\sin }^{4}}\theta \] as \[{{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta \]. So, we will get LHS as

\[LHS=\dfrac{{{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}^{2}}-2{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

\[LHS=\dfrac{{{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}^{2}}-{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

Now, we will again put \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, we will get,

\[LHS=\dfrac{1-{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{2+{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\]

LHS = RHS

Hence proved


Note: In this question, there are lots of possibilities that we might make a calculation mistake. Also, we could have made a mistake while applying the algebraic formulas and trigonometric formulas. So, we have to be very patient while proving the expression.