Question

Prove the following expression:
\[\dfrac{\left( 1+\cot \theta +\tan \theta \right)\left( \sin \theta -\cos \theta \right)}{{{\sec }^{3}}\theta -{{\operatorname{cosec}}^{3}}\theta }={{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

Answer 1

Hint: To prove this expression, we need to know the formulas like how to express tan, cot, sec, and cosec in terms of sin and cos. Also, we should know the identity \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. By using these relations we will be able to prove the expression.

Complete step-by-step answer:

In this question, we are asked to prove that

\[\dfrac{\left( 1+\cot \theta +\tan \theta \right)\left( \sin \theta -\cos \theta \right)}{{{\sec }^{3}}\theta -{{\operatorname{cosec}}^{3}}\theta }={{\sin }^{2}}\theta {{\cos }^{2}}\theta \]

To prove this relation, we will first consider the left-hand side of the expression. So, we get,

\[LHS=\dfrac{\left( 1+\cot \theta +\tan \theta \right)\left( \sin \theta -\cos \theta \right)}{{{\sec }^{3}}\theta -{{\operatorname{cosec}}^{3}}\theta }\]

Now, we will first express all the terms in the terms of sin and cos, that is we will put \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\], \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\],

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

\[LHS=\dfrac{\left( 1+\dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta } \right)\left( \sin \theta -\cos \theta \right)}{\dfrac{1}{{{\cos }^{3}}\theta }-\dfrac{1}{{{\sin }^{3}}\theta }}\]

Now, we will take the LCM of the terms of the numerator and the terms of the denominator. So, we get,

\[LHS=\dfrac{\left[ \dfrac{\sin \theta \cos \theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\sin \theta \cos \theta } \right]\left( \sin \theta -\cos \theta \right)}{\dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta {{\cos }^{3}}\theta }}\]

\[LHS=\dfrac{\left( \sin \theta \cos \theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)\left( \sin \theta -\cos \theta \right)\left( {{\sin }^{3}}\theta {{\cos }^{3}}\theta \right)}{\left( \sin \theta \cos \theta \right)\left( {{\sin }^{3}}\theta -{{\cos }^{3}}\theta \right)}\]

Now, we can see that \[\sin \theta \] and \[\cos \theta \] both are common in numerator and denominator. So, we can cancel them out. Also, we know that \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. Therefore, we will get LHS as

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

Now, we can see that \[\left( \sin \theta -\cos \theta \right)\] and \[\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta \right)\] common in both numerator and denominator. So, we can cancel them out. Therefore, we will get LHS as

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

LHS = RHS

Hence, we have proved the required result.


Note: In this question, one might think to open the brackets of left-hand side and then after multiplication might think to convert tan, cot, sec, and cosec in terms of sin and cos, that is \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\], \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\], \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. But after a few steps, we will come back to the same stage from where we started now. So, it is better to make the solution smaller and easier.