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),

& \cot x\left( 1+\sin x \right)=4m-(1) \\\ & \cot x\left( 1-\sin x \right)=4n-(2) \\\ \end{aligned}$$ Now let us multiply both equations. Multiplying the terms in the LHS we get, $$\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{aligned} & \therefore LHS={{\cot }^{2}}x\left( 1-{{\sin }^{2}}x \right) \\\ & RHS=4m\times 4n=16mn \\\ \end{aligned}$$ Therefore by multiplying both equations we get, $${{\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{aligned} & \dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}\times {{\cos }^{2}}x=16mn \\\ & \therefore \dfrac{{{\cos }^{4}}x}{{{\sin }^{2}}x}=16mn \\\ \end{aligned}$$ So, we get the value of $$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{aligned} & 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{aligned}$$ Cancel out the like terms in LHS of the solution. $$\begin{aligned} & 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{aligned}$$ Let us square on both sides of equation (5). $${{\left( {{m}^{2}}-{{n}^{2}} \right)}^{2}}=\dfrac{\sin x{{\cot }^{2}}x}{16}$$ We know,$$\cot x=\dfrac{\cos x}{\sin x}$$ $$\begin{aligned} & \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{aligned}$$ Now compare equations (4) and (6). $$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.