Question

If ${{\tan }^{2}}x+2\tan x\tan 2y={{\tan }^{2}}y+2\tan y\tan 2x$ then prove that the value of L.H.S and R.H.S is equal to 1 and $\tan x=\pm \tan y$ ?

Answer 1

Let us assume the given equation is equal to k then take one by one L.H.S and R.H.S and equate them equal to k and simplify the equations. Then two equations will be formed which will look like: ${{\tan }^{2}}x+2\tan x\tan 2y=k;{{\tan }^{2}}y+2\tan y\tan 2x=k$. To simplify these two equations, we are going to use the tangent trigonometry identity which is equal to $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$.

Complete step by step solution:
In the above problem, we have given the following equation:
${{\tan }^{2}}x+2\tan x\tan 2y={{\tan }^{2}}y+2\tan y\tan 2x$
Let us equate the above equation to k and we get,
$\Rightarrow {{\tan }^{2}}x+2\tan x\tan 2y={{\tan }^{2}}y+2\tan y\tan 2x=k$
Now, solving the above equations we get,
$\begin{aligned} & {{\tan }^{2}}x+2\tan x\tan 2y=k; \\\ & {{\tan }^{2}}y+2\tan y\tan 2x=k \\\ \end{aligned}$
First of all, we are solving the first equation from the above two equations and we get,
$\Rightarrow {{\tan }^{2}}x+2\tan x\tan 2y=k$
Subtracting ${{\tan }^{2}}x$ on both the sides we get,
$\Rightarrow 2\tan x\tan 2y=k-{{\tan }^{2}}x$
Dividing $2\tan x$ on both the sides of the above equation and we get,
$\Rightarrow \tan 2y=\dfrac{k-{{\tan }^{2}}x}{2\tan x}$
We know that there is a trigonometry identity which states that:
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
Applying the above property in the L.H.S of the above equation and we get,
$\Rightarrow \dfrac{2\tan y}{1-{{\tan }^{2}}y}=\dfrac{k-{{\tan }^{2}}x}{2\tan x}$
Cross multiplying the above equation we get,
$\Rightarrow 4\tan x\tan y=\left( k-{{\tan }^{2}}x \right)\left( 1-{{\tan }^{2}}y \right)$ …………. (1)
Similarly, we are going to equate k with R.H.S of the main equation and we get,
$\Rightarrow {{\tan }^{2}}y+2\tan y\tan 2x=k$
Subtracting ${{\tan }^{2}}y$ on both the sides we get,
$\Rightarrow 2\tan y\tan 2x=k-{{\tan }^{2}}y$
Dividing $2\tan y$ on both the sides of the above equation and we get,
$\Rightarrow \tan 2x=\dfrac{k-{{\tan }^{2}}y}{2\tan y}$
We know that there is a trigonometry identity which states that:
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
Applying the above property in the L.H.S of the above equation and we get,
$\Rightarrow \dfrac{2\tan x}{1-{{\tan }^{2}}x}=\dfrac{k-{{\tan }^{2}}y}{2\tan y}$
Cross multiplying the above equation we get,
$\Rightarrow 4\tan x\tan y=\left( k-{{\tan }^{2}}y \right)\left( 1-{{\tan }^{2}}x \right)$ …………. (2)
From eq. (1) and eq. (2) we get,
$\Rightarrow \left( k-{{\tan }^{2}}x \right)\left( 1-{{\tan }^{2}}y \right)=\left( k-{{\tan }^{2}}y \right)\left( 1-{{\tan }^{2}}x \right)$
Solving the L.H.S and R.H.S of the above equation and we get,
$\Rightarrow k+{{\tan }^{2}}x{{\tan }^{2}}y-k{{\tan }^{2}}y-{{\tan }^{2}}x=k+{{\tan }^{2}}x{{\tan }^{2}}y-k{{\tan }^{2}}x-{{\tan }^{2}}y$
In the above equation, $k\And {{\tan }^{2}}x{{\tan }^{2}}y$ will be cancelled out from both the sides and we get,
$\Rightarrow -k{{\tan }^{2}}y-{{\tan }^{2}}x=-k{{\tan }^{2}}x-{{\tan }^{2}}y$
Rearranging the above equation and we get,
$\begin{aligned} & \Rightarrow k{{\tan }^{2}}x-k{{\tan }^{2}}y+{{\tan }^{2}}y-{{\tan }^{2}}x=0 \\\ & \Rightarrow k\left( {{\tan }^{2}}x-{{\tan }^{2}}y \right)-\left( {{\tan }^{2}}x-{{\tan }^{2}}y \right)=0 \\\ \end{aligned}$
Taking $\left( {{\tan }^{2}}x-{{\tan }^{2}}y \right)$ as common from the L.H.S of the above equation and we get,
$\Rightarrow \left( {{\tan }^{2}}x-{{\tan }^{2}}y \right)\left( k-1 \right)=0$
Equating each bracket to 0 we get,
$\begin{aligned} & \Rightarrow k=1; \\\ & \Rightarrow {{\tan }^{2}}x-{{\tan }^{2}}y=0 \\\ & \Rightarrow \left( \tan x-\tan y \right)\left( \tan x+\tan y \right)=0 \\\ & \Rightarrow \tan x=\pm \tan y \\\ \end{aligned}$

Note: To solve the above problem you must need to know the property of double angle of tangent which is equal to:
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
If you forget this property then you cannot move forward in this problem.