Question

Let ${{\tan }^{2}}x=1+2{{\tan }^{2}}y$, then prove that $\cos 2y=1+2\cos 2x$.

Answer 1

Hint: We try to use the formula of multiple angles relating $\cos 2x$ and $\tan x$. The theorem is $\cos 2y=\dfrac{1-{{\tan }^{2}}y}{1+{{\tan }^{2}}y}$. Using the formula, we find the value of $\cos 2y$ in terms of ${{\tan }^{2}}x$. We also have to break the numerator of the equation to form the relation for $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$.
We replace the values to prove $\cos 2y=1+2\cos 2x$.

Complete step by step answer:
We have been given ${{\tan }^{2}}x=1+2{{\tan }^{2}}y$. We convert the trigonometric ratios using the formulas of multiple angles.
We know that $\cos 2y=\dfrac{1-{{\tan }^{2}}y}{1+{{\tan }^{2}}y}$. So, we find the value of ${{\tan }^{2}}y$ from the equation ${{\tan }^{2}}x=1+2{{\tan }^{2}}y$.
So, ${{\tan }^{2}}x=1+2{{\tan }^{2}}y\Rightarrow {{\tan }^{2}}y=\dfrac{{{\tan }^{2}}x-1}{2}$.
We place the value in the left-hand side of the equation $\cos 2y=1+2\cos 2x$.
$\cos 2y=\dfrac{1-{{\tan }^{2}}y}{1+{{\tan }^{2}}y}=\dfrac{1-\dfrac{{{\tan }^{2}}x-1}{2}}{1+\dfrac{{{\tan }^{2}}x-1}{2}}$.
We need to solve the equation now to find the form of $\cos 2y=1+2\cos 2x$.
We first multiply both denominator and numerator with 2.
$\cos 2y=\dfrac{1-\dfrac{{{\tan }^{2}}x-1}{2}}{1+\dfrac{{{\tan }^{2}}x-1}{2}}=\dfrac{2-\left( {{\tan }^{2}}x-1 \right)}{2+\left( {{\tan }^{2}}x-1 \right)}=\dfrac{3-{{\tan }^{2}}x}{{{\tan }^{2}}x+1}$
Now we take out a constant 1 from the equation.
$\cos 2y=\dfrac{3-{{\tan }^{2}}x}{{{\tan }^{2}}x+1}=\dfrac{{{\tan }^{2}}x+1+2-2{{\tan }^{2}}x}{{{\tan }^{2}}x+1}$
We break it into two parts and use the theorem $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$.
$\cos 2y=\dfrac{{{\tan }^{2}}x+1+2-2{{\tan }^{2}}x}{{{\tan }^{2}}x+1}=\dfrac{{{\tan }^{2}}x+1}{{{\tan }^{2}}x+1}+\dfrac{2\left( 1-{{\tan }^{2}}x \right)}{1+{{\tan }^{2}}x}=1+2\cos 2x$.
Thus proved.

Note: We also can find the derived form of both sides of the to prove statement $\cos 2y=1+2\cos 2x$. We have already got $\cos 2y=\dfrac{3-{{\tan }^{2}}x}{{{\tan }^{2}}x+1}$. Now we try to form the same value for the equation $1+2\cos 2x$ where
$\begin{align}
  & 1+2\cos 2x \\
 & =1+\dfrac{2\left( 1-{{\tan }^{2}}x \right)}{1-{{\tan }^{2}}x} \\
 & =\dfrac{{{\tan }^{2}}x+1+2-2{{\tan }^{2}}x}{{{\tan }^{2}}x+1} \\
 & =\dfrac{3-{{\tan }^{2}}x}{{{\tan }^{2}}x+1} \\
\end{align}$
So, both R.H.S and L.H.S match and the statement is proved.