Question

How do you verify $\sec x\operatorname{cosec}x=\tan x+\cot x$ ?

Answer 1

We can write tan x as $\dfrac{\sin x}{\cos x}$ and we can write cot x as $\dfrac{\cos x}{\sin x}$ and then we can prove the given equation. We know that the sum of square of sin x and square of cos x is equal to 1 and sec x is the reciprocal of cos x, sin x is the reciprocal of cosec x.

Complete step by step answer:
We have to verify $\sec x\operatorname{cosec}x=\tan x+\cot x$
We will start from LHS
We can write tan x as $\dfrac{\sin x}{\cos x}$ and we can write cot x as $\dfrac{\cos x}{\sin x}$
So $\Rightarrow \tan x+\cot x=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}$
Now we add both terms like simple fraction addition
We can write $\Rightarrow \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}$
We know that the value of ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
So we can write $\Rightarrow \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}=\dfrac{1}{\sin x\cos x}$
We know that $\dfrac{1}{\sin x}=\operatorname{cosec}x$ and $\dfrac{1}{\cos x}=\sec x$
So we can write $\Rightarrow \dfrac{1}{\sin x\cos x}=\sec x\operatorname{cosec}x$
So we have proven that $\sec x\operatorname{cosec}x=\tan x+\cot x$ where x is not equal to 0 or $\dfrac{\pi }{2}$ .

Note:
We can see that in the equation given in the question tan x , cosec x , sec x and cot x is
We know that 0 is not include in the domain of cot x and cosec x and $\dfrac{\pi }{2}$ is not included in the domain of sec x and tan x, so while writing the proof we need to exclude 0 and $\dfrac{\pi }{2}$ from the domain of the x. So always keep in mind that it is good practice to mention the domain of the function after writing it. Because we can see from the above example we can write $\sec x\operatorname{cosec}x=\tan x+\cot x$ when x is equal to 0 or $\dfrac{\pi }{2}$