Question

Eliminate x from equation
\[\begin{align}
  & m\tan x+n\cot 2x=p \\
 & m\cot x-n\tan 2x=p \\
\end{align}\]

Answer 1

Hint:Here, we will use various trigonometric identities to eliminate x which are:
\[\begin{align}
  & \left( i \right)\tan x=\dfrac{\sin x}{\cos x} \\
 & \left( ii \right)\cot x=\dfrac{\cos x}{\sin x} \\
 & \left( iii \right){{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \left( iv \right){{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x \\
 & \left( v \right)\sin 2x=2\sin x\cos x \\
 & \left( vi \right){{\sin }^{2}}x=\dfrac{1-\cos 2x}{2} \\
 & \left( vii \right)\sin 2x=\sqrt{1-{{\cos }^{2}}2x} \\
\end{align}\]
We will find an equation in terms of m, n and p by finding value of cos2x from both equations, converting first equation in cos2x form and then eliminating x to find required equation. We will start by subtracting the second equation from the first equation and then apply the above mentioned identities.

Complete step by step answer:
We are given equations:
\[\begin{align}
  & m\tan x+n\cot 2x=p\cdots \cdots \cdots \cdots \left( i \right) \\
 & m\cot x-n\tan 2x=p\cdots \cdots \cdots \cdots \left( ii \right) \\
\end{align}\]
To simplify, subtracting (ii) from (i) we get:
\[\begin{align}
  & m\tan x+n\cot 2x-m\cot x+n\tan x=p-p \\
 & m\left( \cot x-\tan x \right)-n\left( \tan 2x+\cot 2x \right)=0 \\
\end{align}\]
Here, we will use $\tan x=\dfrac{\sin x}{\cos x}\text{ and }\cot x=\dfrac{\cos x}{\sin x}$
\[\begin{align}
  & m\left( \dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x} \right)-n\left( \dfrac{\sin 2x}{\cos 2x}+\dfrac{\cos 2x}{\sin 2x} \right)=0 \\
 & \Rightarrow m\left( \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\sin x\cos x} \right)=n\left( \dfrac{{{\sin }^{2}}2x+{{\cos }^{2}}2x}{\sin 2x\cos 2x} \right) \\
\end{align}\]
Using ${{\sin }^{2}}x+{{\cos }^{2}}x=1\text{ and }{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
Also using $\sin 2x=2\sin x\cos x$ we get:
\[\begin{align}
  & m\left( \dfrac{\cos 2x}{\sin x\cos x} \right)=n\left( \dfrac{1}{2\sin x\cos x\cos 2x} \right) \\
 & \Rightarrow m\left( \dfrac{2{{\cos }^{2}}2x\sin x\cos x}{\sin x\cos x} \right)=n \\
\end{align}\]
Eliminating sinx cosx from numerator and denominator, we get:
\[\begin{align}
  & 2m{{\cos }^{2}}2x=n \\
 & \Rightarrow {{\cos }^{2}}2x=\dfrac{n}{2m} \\
\end{align}\]
Taking square root both sides, we get:
\[\cos 2x=\sqrt{\dfrac{n}{2m}}\cdots \cdots \cdots \cdots \left( iii \right)\]
We have found value of cos2x, now let us convert equation (i) in terms of cos2x.
Taking equation (i) \[m\tan x+n\cot 2x=p\]
Using $\tan x=\dfrac{\sin x}{\cos x}\text{ and }\cot 2x=\dfrac{\cos 2x}{\sin 2x}$ we get:
\[m\dfrac{\sin x}{\cos x}+n\dfrac{\cos 2x}{\sin 2x}=p\]
Taking LCM we get:
\[\dfrac{m\sin x\sin 2x+n\cos 2x\cos x}{\cos x\sin 2x}=p\]
Using $\sin 2x=2\sin x\cos x$ we get:
\[\dfrac{2m{{\sin }^{2}}x\cos x+n\cos 2x\cos x}{\cos x\sin 2x}=p\]
Eliminating cosx from numerator and denominator, we get:
\[\begin{align}
  & \dfrac{2m{{\sin }^{2}}x+n\cos 2x}{\sin 2x}=p \\
 & \Rightarrow 2m{{\sin }^{2}}x+n\cos 2x=p\sin 2x \\
\end{align}\]
Using ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}\text{ and }\sin 2x=\sqrt{1-{{\cos }^{2}}2x}$ we get:
\[2m\left( \dfrac{1-\cos 2x}{2} \right)+n\cos 2x=p\sqrt{1-{{\cos }^{2}}2x}\]
As we can see whole equation is converted in terms of cos2x, we can put value of cos2x in above equation to find our required equation.
Putting value of cos2x from (iii) in above equation, we get:
\[m\left( 1-\sqrt{\dfrac{n}{2m}} \right)+n\left( \sqrt{\dfrac{n}{2m}} \right)=p\sqrt{1-\dfrac{n}{2m}}\]
Simplifying above equation, we get:
\[m-\left( m-n \right)\sqrt{\dfrac{n}{2m}}=p\sqrt{2m-n}\]
Removing denominator term to simplify equation more, we get:
\[m\sqrt{2m-m+n}\sqrt{n}=p\sqrt{2m-n}\]
As we can see equation is purely in terms of m, n and p, hence, we have eliminated x.

Note:
 Here, students can further simplify the equation to obtain required answer. Students have to remember all the trigonometric identities. While subtracting the equations, take care of calculation. Select only the equation which can be converted in a single trigonometric function to reach answer.