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Question: How do you verify \(\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x\) ?...

How do you verify 1−tan⁡2x1+tan⁡2x=1−2sin⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x ?

Explanation

Solution

We know the formula 1+tan⁡2x1+{{\tan }^{2}}x is equal to sec⁡2x{{\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 1−tan⁡2x1+tan⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} is equal to 1−2sin⁡2x1-2{{\sin }^{2}}x. We replace tan x with sin⁡xcos⁡x\dfrac{\sin x}{\cos x} and 1+tan⁡2x1+{{\tan }^{2}}x with sec⁡2x{{\sec }^{2}}x .

Complete step-by-step answer:
We have to verify 1−tan⁡2x1+tan⁡2x=1−2sin⁡2x\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⁡2x1+{{\tan }^{2}}x = sec⁡2x{{\sec }^{2}}x and tan x = sin⁡xcos⁡x\dfrac{\sin x}{\cos x} , we can write tan⁡2x{{\tan }^{2}}x equal to sin⁡2xcos⁡2x\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}
So we can write 1−tan⁡2x1+tan⁡2x=1−sin⁡2xcos⁡2x1+sin⁡2xcos⁡2x\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 nπ2\dfrac{n\pi }{2} where n is an integer.
Further solving we get 1−tan⁡2x1+tan⁡2x=cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x\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⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
⇒1−tan⁡2x1+tan⁡2x=cos⁡2x−sin⁡2x\Rightarrow \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}={{\cos }^{2}}x-{{\sin }^{2}}x
Now replacing cos⁡2x{{\cos }^{2}}x with 1−sin⁡2x1-{{\sin }^{2}}x in the above equation we get
⇒1−tan⁡2x1+tan⁡2x=1−2sin⁡2x\Rightarrow \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=1-2{{\sin }^{2}}x where x is not equal to nπ2\dfrac{n\pi }{2}

Note: We can prove the value of 1−tan⁡2x1+tan⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} and 1−2sin⁡2x1-2{{\sin }^{2}}x both are equal to cos 2x . Another formula for cos 2x is 2cos⁡2x−12{{\cos }^{2}}x-1 . While we write cos 2x is equal to 1−tan⁡2x1+tan⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} , we should always mention that the angle x should be equal to nπ2\dfrac{n\pi }{2} ; where n is an integer, because at nπ2\dfrac{n\pi }{2} the value of tan x tends to infinity. nπ2\dfrac{n\pi }{2} does not lie in the domain of tan x.