Question

If $\text{cosecx}-\text{sinx}={{a}^{3}}$ and $\sec x-\cos x={{b}^{3}}$, prove that ${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})$ = 1.

Answer 1

Hint:To solve this question, it is important to be aware about the basic concepts related to trigonometric equations. In this question, we will first try to convert cosec(x) into sin(x) and sec(x) into cos(x). We would then simplify the terms as per ${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})$ term to get the necessary answer.

Complete step-by-step answer:
The first step of the problem would be to simplify $\text{cscx}-\text{sinx}={{a}^{3}}$ further. Thus, first we convert cosec(x) into sin(x). Then, we get,
$\begin{align}
  & \dfrac{1}{\sin x}-\sin x={{a}^{3}} \\
 & \dfrac{1-{{\sin }^{2}}x}{\sin x}={{a}^{3}} \\
 & \dfrac{{{\cos }^{2}}x}{\sin x}={{a}^{3}} \\
\end{align}$
Now, we similarly repeat this for the term $\sec x-\cos x={{b}^{3}}$, this time we convert sec(x) into cos(x). Now, we get,
$\begin{align}
  & \dfrac{1}{\cos x}-\cos x={{b}^{3}} \\
 & \dfrac{1-{{\cos }^{2}}x}{\cos x}={{b}^{3}} \\
 & \dfrac{{{\sin }^{2}}x}{\cos x}={{b}^{3}} \\
\end{align}$
Now, we multiply the obtained results, thus, we get,
$\left( \dfrac{{{\cos }^{2}}x}{\sin x} \right)\left( \dfrac{{{\sin }^{2}}x}{\cos x} \right)={{a}^{3}}{{b}^{3}}$
cos(x)sin(x) = ${{a}^{3}}{{b}^{3}}$ -- (1)
Now, we observe that we want to evaluate ${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})$. Thus, we first find ${{a}^{2}}{{b}^{2}}$.
Thus, in equation (1), we raise LHS and RHS to the power of $\dfrac{2}{3}$. Doing so, we get,
${{\left( \cos x\sin x \right)}^{\dfrac{2}{3}}}={{a}^{2}}{{b}^{2}}$ -- (2)
Next, we proceed to find ${{a}^{2}}+{{b}^{2}}$. For this, we make use of the fact that
$\begin{align}
  & {{a}^{3}}=\left( \dfrac{{{\cos }^{2}}x}{\sin x} \right) \\
 & {{b}^{3}}=\left( \dfrac{{{\sin }^{2}}x}{\cos x} \right) \\
\end{align}$
Thus, \[\begin{align}
  & {{a}^{2}}={{\left( \dfrac{{{\cos }^{2}}x}{\sin x} \right)}^{\dfrac{2}{3}}}=\left( \dfrac{{{\cos }^{\dfrac{4}{3}}}x}{{{\sin }^{\dfrac{2}{3}}}x} \right) \\
 & {{b}^{2}}={{\left( \dfrac{{{\sin }^{2}}x}{\cos x} \right)}^{\dfrac{2}{3}}}=\left( \dfrac{{{\sin }^{\dfrac{4}{3}}}x}{{{\cos }^{\dfrac{2}{3}}}x} \right) \\
\end{align}\]
Now, we add these results, we get,
${{a}^{2}}+{{b}^{2}}=\left( \dfrac{{{\cos }^{\dfrac{4}{3}}}x}{{{\sin }^{\dfrac{2}{3}}}x} \right)+\left( \dfrac{{{\sin }^{\dfrac{4}{3}}}x}{{{\cos }^{\dfrac{2}{3}}}x} \right)$
${{a}^{2}}+{{b}^{2}}=\left( \dfrac{{{\cos }^{\dfrac{4}{3}}}x{{\cos }^{\dfrac{2}{3}}}x+{{\sin }^{\dfrac{4}{3}}}x{{\sin }^{\dfrac{2}{3}}}x}{{{\sin }^{\dfrac{2}{3}}}x{{\cos }^{\dfrac{2}{3}}}x} \right)$
${{a}^{2}}+{{b}^{2}}=\left( \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\sin }^{\dfrac{2}{3}}}x{{\cos }^{\dfrac{2}{3}}}x} \right)$
Now, we use the property that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Thus, we get,
${{a}^{2}}+{{b}^{2}}=\left( \dfrac{1}{{{\sin }^{\dfrac{2}{3}}}x{{\cos }^{\dfrac{2}{3}}}x} \right)$ -- (3)
Now, we combine the results from (2) and (3). Thus, we get,
\[\left( {{a}^{2}}{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)=\left( \dfrac{{{\left( \cos x\sin x \right)}^{\dfrac{2}{3}}}}{{{\sin }^{\dfrac{2}{3}}}x{{\cos }^{\dfrac{2}{3}}}x} \right)\]
Since, numerator and the denominator are the same, we get,
\[\left( {{a}^{2}}{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)=1\]
Hence, LHS=RHS, we have proved the above problem.

Note: In this problem, before jumping straight away to evaluate the term ${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})$, first simplifying the expression in hand helps in solving the problem with ease. One can also try to solve${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})$ term directly without simplifying $\text{cosec x}-\sin x={{a}^{3}}$ and $\sec x-\cos x={{b}^{3}}$ terms, however, this would be an extremely tedious task and it may also happen that problem is unsolvable by this method.