Question

How do you verify $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x$ ?

Answer 1

Hint: We know the formula $1+{{\tan }^{2}}x$ is equal to ${{\sec }^{2}}x$ , sum of squares of sin x and cos x is equal to 1 and we know that tan x is the ratio of sin x and cos x . We can use these 2 formula to prove $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ is equal to $1-2{{\sin }^{2}}x$. We replace tan x with $\dfrac{\sin x}{\cos x}$ and $1+{{\tan }^{2}}x$ with ${{\sec }^{2}}x$ .

Complete step-by-step answer:
We have to verify $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x$ , we will go from LHS to RHS
 we know that $1+{{\tan }^{2}}x$ = ${{\sec }^{2}}x$ and tan x = $\dfrac{\sin x}{\cos x}$ , we can write ${{\tan }^{2}}x$ equal to $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$
So we can write $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\dfrac{1-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}$ where cos x is not equal to 0 that means x is not equal to $\dfrac{n\pi }{2}$ where n is an integer.
Further solving we get $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x}$ where cos x is not equal to 0
We know that sum of squares of sin x and cos x is equal to 1 so we can write ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\Rightarrow \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}={{\cos }^{2}}x-{{\sin }^{2}}x$
Now replacing ${{\cos }^{2}}x$ with $1-{{\sin }^{2}}x$ in the above equation we get
$\Rightarrow \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x$ where x is not equal to $\dfrac{n\pi }{2}$

Note: We can prove the value of $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and $1-2{{\sin }^{2}}x$ both are equal to cos 2x . Another formula for cos 2x is $2{{\cos }^{2}}x-1$ . While we write cos 2x is equal to $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ , we should always mention that the angle x should be equal to $\dfrac{n\pi }{2}$ ; where n is an integer, because at $\dfrac{n\pi }{2}$ the value of tan x tends to infinity. $\dfrac{n\pi }{2}$ does not lie in the domain of tan x.