Question

If trigonometric identiites $\csc \theta -\sin \theta ={{b}^{3}}$ and $\sec \theta -\cos \theta ={{a}^{3}}$ then prove that ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)=1$ .

Answer 1

Hint: For solving this question first we will simplify the value of $a$ and $b$ to find their values. We will use one of the very basic trigonometric identities and that is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . Moreover, we will perform some simple algebraic operations and prove the desired result.

Complete step-by-step solution -
Given:
It is given that if $\csc \theta -\sin \theta ={{b}^{3}}$ and $\sec \theta -\cos \theta ={{a}^{3}}$ . We have to prove that ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)=1$ .
Now, before we proceed we should know the following equations:
$\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta ......................\left( 1 \right) \\
 & \Rightarrow 1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta ......................\left( 2 \right) \\
 & \dfrac{1}{\sin \theta }=\csc \theta ...................................\left( 3 \right) \\
 & \dfrac{1}{\cos \theta }=\sec \theta ..................................\left( 4 \right) \\
\end{align}$
Now, we will use the formulas in the above equations to simplify the value of $a$ , $b$ and then we will prove the desired result.
Simplification of $b$ :
Now, we have $\csc \theta -\sin \theta ={{b}^{3}}$ so, using the formula from the equation (3). Then,
$\begin{align}
  & \csc \theta -\sin \theta ={{b}^{3}} \\
 & \Rightarrow \dfrac{1}{\sin \theta }-\sin \theta ={{b}^{3}} \\
 & \Rightarrow {{b}^{3}}=\dfrac{1}{\sin \theta }-\sin \theta \\
 & \Rightarrow {{b}^{3}}=\dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \\
\end{align}$
Now, using the formula form equation (2) in the above equation. Then,
${{b}^{3}}=\dfrac{1-{{\sin }^{2}}\theta }{\sin \theta }$
$\begin{align}
  & \Rightarrow {{b}^{3}}=\dfrac{{{\cos }^{2}}\theta }{\sin \theta } \\
 & \Rightarrow b=\dfrac{{{\cos }^{{}^{2}/{}_{3}}}\theta }{{{\sin }^{{}^{1}/{}_{3}}}\theta }........................\left( 5 \right) \\
\end{align}$
Simplification of $a$ :
Now, we have $\sec \theta -\cos \theta ={{a}^{3}}$ so, using the formula from the equation (4). Then,
$\begin{align}
  & \sec \theta -\cos \theta ={{a}^{3}} \\
 & \Rightarrow \dfrac{1}{\cos \theta }-\cos \theta ={{a}^{3}} \\
 & \Rightarrow \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }={{a}^{3}} \\
 & \Rightarrow {{a}^{3}}=\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \\
\end{align}$
Now, using the formula form equation (1) in the above equation. Then,
$\begin{align}
  & {{a}^{3}}=\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \\
 & \Rightarrow {{a}^{3}}=\dfrac{{{\sin }^{2}}\theta }{\cos \theta } \\
 & \Rightarrow a=\dfrac{{{\sin }^{{}^{2}/{}_{3}}}\theta }{{{\cos }^{{}^{1}/{}_{3}}}\theta }..............................\left( 5 \right) \\
\end{align}$
Now, from the equation (4) and equation (5) substituting the value of $a$ and $b$ in the expression ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$ . Then,
\[\begin{align}
  & {{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right) \\
 & \Rightarrow {{\left( \dfrac{{{\sin }^{{}^{2}/{}_{3}}}\theta }{{{\cos }^{{}^{1}/{}_{3}}}\theta }\times \dfrac{{{\cos }^{{}^{2}/{}_{3}}}\theta }{{{\sin }^{{}^{1}/{}_{3}}}\theta } \right)}^{2}}\times \left( {{\left( \dfrac{{{\sin }^{{}^{2}/{}_{3}}}\theta }{{{\cos }^{{}^{1}/{}_{3}}}\theta } \right)}^{2}}+{{\left( \dfrac{{{\cos }^{{}^{2}/{}_{3}}}\theta }{{{\sin }^{{}^{1}/{}_{3}}}\theta } \right)}^{2}} \right) \\
 & \Rightarrow {{\left( {{\sin }^{{}^{1}/{}_{3}}}\theta {{\cos }^{{}^{1}/{}_{3}}}\theta \right)}^{2}}\times \left( \dfrac{{{\sin }^{{}^{4}/{}_{3}}}\theta }{{{\cos }^{{}^{2}/{}_{3}}}\theta }+\dfrac{{{\cos }^{{}^{4}/{}_{3}}}\theta }{{{\sin }^{{}^{2}/{}_{3}}}\theta } \right) \\
 & \Rightarrow {{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta \times \left( \dfrac{{{\sin }^{{}^{6}/{}_{3}}}\theta +{{\cos }^{{}^{6}/{}_{3}}}\theta }{{{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta } \right) \\
 & \Rightarrow {{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta \times \left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta } \right) \\
\end{align}\]


\[\begin{align}
  & \Rightarrow {{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta \times \dfrac{1}{{{\sin }^{{}^{2}/{}_{3}}}\theta {{\cos }^{{}^{2}/{}_{3}}}\theta } \\
 & \Rightarrow 1 \\
\end{align}\]
Now, from the above result, we can say that the value of ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$ is equal to 1.
Thus, ${{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)=1$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question and try to apply the perfect formula. Moreover, the student should proceed in a stepwise manner and while solving first find the value of $a$ and $b$ separately and then prove the result comfortably.