Courses
Courses for Kids
Free study material
Free LIVE classes
More

# 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
Total views: 304.8k
Views today: 6.85k
Verified
304.8k+ views
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 ${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn$.

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.