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

Last updated date: 18th Mar 2023

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Answer

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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.

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.

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