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Question: How do you prove \(2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \righ...

How do you prove 2tan⁡xsec⁡x=(11−sin⁡x)−(11+sin⁡x)2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right) ?

Explanation

Solution

We know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and we know that sum of square of sin x and cos x 1 we can use this this prove 2tan⁡xsec⁡x=(11−sin⁡x)−(11+sin⁡x)2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right) and we have to keep in mind that sin x can not be equal to 1 or -1.

Complete step-by-step answer:
We have to prove 2tan⁡xsec⁡x=(11−sin⁡x)−(11+sin⁡x)2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right) , we will prove from RHS to LHS
So we have (11−sin⁡x)−(11+sin⁡x)\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)
Taking LCM of denominator and solve the above equation we get
⇒(11−sin⁡x)−(11+sin⁡x)=1+sin⁡x−1+sin⁡x(1−sin⁡x)(1+sin⁡x)\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{1+\sin x-1+\sin x}{\left( 1-\sin x \right)\left( 1+\sin x \right)}
We know the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} we can apply it
⇒(11−sin⁡x)−(11+sin⁡x)=2sin⁡x(1−sin⁡2x)\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{\left( 1-{{\sin }^{2}}x \right)}
We know that the value of (1−sin⁡2x)\left( 1-{{\sin }^{2}}x \right) is equal to cos⁡2x{{\cos }^{2}}x
So we can write (11−sin⁡x)−(11+sin⁡x)=2sin⁡xcos⁡2x\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{{{\cos }^{2}}x}
⇒(11−sin⁡x)−(11+sin⁡x)=2×sin⁡xcos⁡x×1cos⁡x\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\times \dfrac{\sin x}{\cos x}\times \dfrac{1}{\cos x}
We know that tan x is the ratio of sin x and cos x and sec x is the reciprocal of cos x
⇒(11−sin⁡x)−(11+sin⁡x)=2tan⁡xsec⁡x\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\tan x\sec x where sin x is not equal to -1 or 1 so x is not equal to nπ2\dfrac{n\pi }{2} where n is an integer.

Note: In the above proof we excluded nπ2\dfrac{n\pi }{2} from the domain , so x can not be equal to nπ2\dfrac{n\pi }{2} because tan x and sec x does not exist at nπ2\dfrac{n\pi }{2} and the denominator will be 0 if we do so. In nπ2\dfrac{n\pi }{2}
n is always an integer we can not put a decimal number in place of x. Always remember the formula for all trigonometric functions such as the sum of squares of sin x and cos x is equal to 1. cos x and sec x , sin x and cosec x , tan x and cot x are reciprocal of each other.