How do you prove $2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$ ?
Answer
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Hint: We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and we know that sum of square of sin x and cos x 1 we can use this this prove $2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$ 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 $2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$ , we will prove from RHS to LHS
So we have $\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$
Taking LCM of denominator and solve the above equation we get
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{1+\sin x-1+\sin x}{\left( 1-\sin x \right)\left( 1+\sin x \right)}$
We know the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we can apply it
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{\left( 1-{{\sin }^{2}}x \right)}$
We know that the value of $\left( 1-{{\sin }^{2}}x \right)$ is equal to ${{\cos }^{2}}x$
So we can write $\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{{{\cos }^{2}}x}$
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\times \dfrac{\sin x}{\cos x}\times \dfrac{1}{\cos x}$
We know that tan x is the ratio of sin x and cos x and sec x is the reciprocal of cos x
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\tan x\sec x$ where sin x is not equal to -1 or 1 so x is not equal to $\dfrac{n\pi }{2}$ where n is an integer.
Note: In the above proof we excluded $\dfrac{n\pi }{2}$ from the domain , so x can not be equal to $\dfrac{n\pi }{2}$ because tan x and sec x does not exist at $\dfrac{n\pi }{2}$ and the denominator will be 0 if we do so. In $\dfrac{n\pi }{2}$
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.
Complete step-by-step answer:
We have to prove $2\tan x\sec x=\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$ , we will prove from RHS to LHS
So we have $\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)$
Taking LCM of denominator and solve the above equation we get
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{1+\sin x-1+\sin x}{\left( 1-\sin x \right)\left( 1+\sin x \right)}$
We know the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ we can apply it
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{\left( 1-{{\sin }^{2}}x \right)}$
We know that the value of $\left( 1-{{\sin }^{2}}x \right)$ is equal to ${{\cos }^{2}}x$
So we can write $\left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=\dfrac{2\sin x}{{{\cos }^{2}}x}$
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\times \dfrac{\sin x}{\cos x}\times \dfrac{1}{\cos x}$
We know that tan x is the ratio of sin x and cos x and sec x is the reciprocal of cos x
$\Rightarrow \left( \dfrac{1}{1-\sin x} \right)-\left( \dfrac{1}{1+\sin x} \right)=2\tan x\sec x$ where sin x is not equal to -1 or 1 so x is not equal to $\dfrac{n\pi }{2}$ where n is an integer.
Note: In the above proof we excluded $\dfrac{n\pi }{2}$ from the domain , so x can not be equal to $\dfrac{n\pi }{2}$ because tan x and sec x does not exist at $\dfrac{n\pi }{2}$ and the denominator will be 0 if we do so. In $\dfrac{n\pi }{2}$
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.
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