Question

If \[\cot x\left( 1+\sin x \right)=4m\] and \[\cot x\left( 1-\sin x \right)=4n\], prove that \[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn\].

Answer 1

Hint: First multiply both the equations to form an equation related to RS of what we have to prove. We get an expression for \[mn\]. Secondly, to square and subtract both equations, we get an expression of \[{{m}^{2}}-{{n}^{2}}\]. Square it and prove that \[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn\].

Complete step-by-step answer:
Here we are given two expressions, make them as equation (1) and equation (2),
\[\begin{align}
 & \cot x\left( 1+\sin x \right)=4m-(1) \\
 & \cot x\left( 1-\sin x \right)=4n-(2) \\
\end{align}\]
Now let us multiply both equations.
Multiplying the terms in the LHS we get, \[\left[ \cot x\left( 1+\sin x \right) \right]\left[ \cot x\left( 1-\sin x \right) \right]\]
\[={{\cot }^{2}}x\left( 1+\sin x \right)\left( 1-\sin x \right)\]
We know, \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
Similarly, if a = 1 and \[b=\sin x\],
\[\left( 1-\sin x \right)\left( 1+\sin x \right)\]becomes\[\Rightarrow 1-{{\sin }^{2}}x\].
\[\begin{align}
 & \therefore LHS={{\cot }^{2}}x\left( 1-{{\sin }^{2}}x \right) \\
 & RHS=4m\times 4n=16mn \\
\end{align}\]

Therefore by multiplying both equations we get,
\[{{\cot }^{2}}x\left( 1-{{\sin }^{2}}x \right)=16mn-(3)\]
We know that, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\therefore {{\cos }^{2}}x=1-{{\sin }^{2}}x\].
Let us substitute \[{{\cos }^{2}}x\]in the place of \[\left( 1-{{\sin }^{2}}x \right)\]in equation (3).
So, equation (3) becomes,
\[{{\cot }^{2}}x{{\cos }^{2}}x=16mn\]
The value of \[\cot x=\dfrac{\cos x}{\sin x}\].
Substituting the value of \[\cot x\], the equation changes to,
\[\begin{align}
 & \dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}\times {{\cos }^{2}}x=16mn \\
 & \therefore \dfrac{{{\cos }^{4}}x}{{{\sin }^{2}}x}=16mn \\
\end{align}\]
So, we get the value of \[mn=\dfrac{{{\cos }^{4}}x}{16{{\sin }^{2}}x}-(4)\]
Now let us square equation (1) and equation (2).
Squaring of equation (1) \[\Rightarrow {{\left( 4m \right)}^{2}}={{\cot }^{2}}x{{\left( 1+\sin x \right)}^{2}}\]
                                                  \[\Rightarrow 16{{m}^{2}}={{\cot }^{2}}x{{\left( 1+\sin x \right)}^{2}}\]
Squaring of equation (2) \[\Rightarrow {{\left( 4n \right)}^{2}}={{\cot }^{2}}x{{\left( 1-\sin x \right)}^{2}}\]
                                             \[\Rightarrow 16{{n}^{2}}={{\cot }^{2}}x{{\left( 1-\sin x \right)}^{2}}\]
Now, let us subtract both these squared equations, we will get
\[16{{m}^{2}}-16{{n}^{2}}=\left[ {{\cot }^{2}}x{{\left( 1+\sin x \right)}^{2}} \right]-\left[ {{\cot }^{2}}x{{\left( 1-\sin x \right)}^{2}} \right]\]
Expand \[{{\left( 1+\sin x \right)}^{2}}\]and \[{{\left( 1-\sin x \right)}^{2}}\].
\[\begin{align}
 & 16\left( {{m}^{2}}-{{n}^{2}} \right)=\left[ {{\cot }^{2}}x\left( 1+2\sin x+{{\sin }^{2}}x \right) \right]-\left[ {{\cot }^{2}}x\left( 1-2\sin x+{{\sin }^{2}}x \right) \right] \\
 & 16\left( {{m}^{2}}-{{n}^{2}} \right)={{\cot }^{2}}x+2\sin x{{\cot }^{2}}x+{{\sin }^{2}}x{{\cot }^{2}}x-{{\cot }^{2}}x+2\sin x{{\cot }^{2}}x-{{\cot }^{2}}x{{\sin }^{2}}x \\
\end{align}\]
Cancel out the like terms in LHS of the solution.
\[\begin{align}
 & 16\left( {{m}^{2}}-{{n}^{2}} \right)=2\sin x{{\cot }^{2}}x+2\sin x{{\cot }^{2}}x \\
 & 16\left( {{m}^{2}}-{{n}^{2}} \right)=4\sin x{{\cot }^{2}}x \\
 & {{m}^{2}}-{{n}^{2}}=\dfrac{\sin x{{\cot }^{2}}x}{16}-(5) \\
\end{align}\]

Let us square on both sides of equation (5).
\[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{\sin x{{\cot }^{2}}x}{16}\]
We know,\[\cot x=\dfrac{\cos x}{\sin x}\]
\[\begin{align}
 & \therefore {{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\sin }^{2}}x\times {{\cos }^{4}}x}{16\times {{\sin }^{4}}x} \\
 & \therefore {{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\cos }^{4}}x}{16{{\sin }^{2}}x}-(6) \\
\end{align}\]

Now compare equations (4) and (6).
\[mn=\dfrac{{{\cos }^{4}}x}{16{{\sin }^{2}}x}\]and \[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{{{\cos }^{4}}x}{16{{\sin }^{2}}x}\].
\[\therefore {{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn\].
Hence proved.

Note: The solution is a bit lengthy, so don’t miss out steps. Remember the basic trigonometric formulas like the value of \[\cot x\], which is used in a lot of places. Don’t confuse between the variables m and n. Be alert while solving problems with more variables. First find an expression for mn by multiplying. Then find the expression for \[{{m}^{2}}-{{n}^{2}}\], by squaring and subtracting the equations.