Question

If we have the expressions, $\csc \theta -\sin \theta ={{a}^{3}},\sec \theta -\cos \theta ={{b}^{3}}$, then prove that, ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)=1$.

Answer 1

Hint: We will use the trigonometric relations, $\csc \theta =\dfrac{1}{\sin \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$ to simplify the given expression. We will then use the relations, $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ and $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $. Finally we will substitute the obtained values $a$ and $b$ in the expression that we have to prove, ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$ to get 1 as the final result.

Complete step-by-step answer:

It is given in the question that $\csc \theta -\sin \theta ={{a}^{3}},\sec \theta -\cos \theta ={{b}^{3}}$, and we have to prove that, ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)=1$.

Let us consider the expression, $\csc \theta -\sin \theta ={{a}^{3}}\ldots \ldots \ldots \left( i \right)$

We know that $\csc \theta =\dfrac{1}{\sin \theta }$, so by substituting the value of $\csc \theta $ in equation (i), we get,

$\dfrac{1}{\sin \theta }-\sin \theta ={{a}^{3}}\ldots \ldots \ldots \left( ii \right)$

By taking the LCM in equation (ii), we will get,

$\dfrac{1-{{\sin }^{2}}\theta }{\sin \theta }={{a}^{3}}\ldots \ldots \ldots \left( iii \right)$

We know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $, so by substituting it equation (iii), we get,

$\begin{align}

  & \dfrac{{{\cos }^{2}}\theta }{\sin \theta }={{a}^{3}} \\

 & \Rightarrow a={{\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{\dfrac{1}{3}}} \\

 & \Rightarrow a=\dfrac{{{\cos }^{\dfrac{2}{3}}}\theta }{{{\sin }^{\dfrac{1}{3}}}\theta } \\

\end{align}$

Similarly, we will now consider the other expression, $\sec \theta -\cos \theta ={{b}^{3}}\ldots \ldots \ldots \left( iv \right)$

We know that $\sec \theta =\dfrac{1}{\cos \theta }$, so by substituting the value of $\sec \theta $ in equation (iv), we get,

$\dfrac{1}{\cos \theta }-\cos \theta ={{b}^{3}}\ldots \ldots \ldots \left( v \right)$

By taking the LCM in equation (v), we will get,

$\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }={{b}^{3}}\ldots \ldots \ldots \left( vi \right)$

We know that $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $, so by substituting it equation (vi),

we get,

$\begin{align}

  & \dfrac{{{\sin }^{2}}\theta }{\cos \theta }={{b}^{3}} \\

 & \Rightarrow b={{\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{\dfrac{1}{3}}} \\

 & \Rightarrow b=\dfrac{{{\sin }^{\dfrac{2}{3}}}\theta }{{{\cos }^{\dfrac{1}{3}}}\theta } \\

\end{align}$

Now, we have, $a=\dfrac{{{\cos }^{\dfrac{2}{3}}}\theta }{{{\sin }^{\dfrac{1}{3}}}\theta }$ and $b=\dfrac{{{\sin }^{\dfrac{2}{3}}}\theta }{{{\cos }^{\dfrac{1}{3}}}\theta }$. We have the left hand side or the LHS of the expression to be proved as, ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$. By substituting the values of $a$ and $b$in this expression, we get,

$\begin{align}

  & {{\left( \dfrac{{{\cos }^{\dfrac{2}{3}}}\theta }{{{\sin }^{\dfrac{1}{3}}}\theta } \right)}^{2}}{{\left( \dfrac{{{\sin }^{\dfrac{2}{3}}}\theta }{{{\cos }^{\dfrac{1}{3}}}\theta } \right)}^{2}}\left\{ {{\left( \dfrac{{{\cos }^{\dfrac{2}{3}}}\theta }{{{\sin }^{\dfrac{1}{3}}}\theta } \right)}^{2}}+{{\left( \dfrac{{{\sin }^{\dfrac{2}{3}}}\theta }{{{\cos }^{\dfrac{1}{3}}}\theta } \right)}^{2}} \right\} \\

 & =\dfrac{{{\cos }^{\dfrac{4}{3}}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta }\times \dfrac{{{\sin }^{\dfrac{4}{3}}}\theta }{{{\cos }^{\dfrac{2}{3}}}\theta }\times \left\{ \dfrac{{{\cos }^{\dfrac{4}{3}}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta }+\dfrac{{{\sin }^{\dfrac{4}{3}}}\theta }{{{\cos }^{\dfrac{2}{3}}}\theta } \right\} \\

 & ={{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta \times \left\{ \dfrac{{{\cos }^{\dfrac{4}{3}}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta }+\dfrac{{{\sin }^{\dfrac{4}{3}}}\theta }{{{\cos }^{\dfrac{2}{3}}}\theta } \right\} \\

 & ={{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta \times \left\{ \dfrac{{{\cos }^{\dfrac{4}{3}}}\theta \times {{\cos }^{\dfrac{2}{3}}}\theta +{{\sin }^{\dfrac{4}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta \times {{\cos }^{\dfrac{2}{3}}}\theta } \right\} \\

 & ={{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta \times \left\{ \dfrac{{{\cos }^{\dfrac{6}{3}}}\theta +{{\sin }^{\dfrac{6}{3}}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta \times {{\cos }^{\dfrac{2}{3}}}\theta } \right\} \\

 & ={{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta \times \left\{ \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{\sin }^{\dfrac{2}{3}}}\theta \times {{\cos }^{\dfrac{2}{3}}}\theta } \right\} \\

\end{align}$

We know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, so by substituting that in the above expression, we get,

${{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta \times \left\{ \dfrac{1}{{{\cos }^{\dfrac{2}{3}}}\theta \times {{\sin }^{\dfrac{2}{3}}}\theta } \right\}$

By cancelling the similar terms, we get, 1, which is the right hand side or the RHS of the given expression. Therefore, LHS = RHS.

Hence, we have proved the expression given in the question.


Note: This is a very basic trigonometric function, but the students usually make the mistakes while doing the operations of multiplication and division of powers in exponents. Hence it is advisable to solve this question step by step and not in one go. In questions, where you have to prove, always pay attention to both sides of the expression, that is the LHS and the RHS to get a better idea of the steps to prove them equal.