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
Answer
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 \[{{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=mn\].
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
