Question

How do you verify $\text{cosec}x-\sin x=\cos x\cot x$?

Answer 1

We use the trigonometric identities like $\text{cosec}x=\dfrac{1}{\sin x}$, $1-{{\sin }^{2}}x={{\cos }^{2}}x$ to convert the given equation of $\text{cosec}x-\sin x$ into a multiplication form. We also use the ratio relation of $\dfrac{\cos x}{\sin x}=\cot x$ to find the final result after those binary operations.

Complete step by step answer:
We need to verify the given equation $\text{cosec}x-\sin x=\cos x\cot x$.
We are going to use the relation between the ratios to prove the equation.
We know that $\text{cosec}x=\dfrac{1}{\sin x}$. Replacing the value in the left-hand side of $\text{cosec}x-\sin x$
$\text{cosec}x-\sin x=\dfrac{1}{\sin x}-\sin x$. Now we simplify the subtraction.
We also know that $1-{{\sin }^{2}}x={{\cos }^{2}}x$. The L.C.M becomes $\sin x$.
Therefore, $\dfrac{1}{\sin x}-\sin x=\dfrac{1-{{\sin }^{2}}x}{\sin x}=\dfrac{{{\cos }^{2}}x}{\sin x}$. We replaced the value of $1-{{\sin }^{2}}x$ with ${{\cos }^{2}}x$.
Now we break the division of $\dfrac{{{\cos }^{2}}x}{\sin x}$ as $\dfrac{\cos x}{\sin x}\times \cos x$.
We have the identity of $\dfrac{\cos x}{\sin x}=\cot x$. We replace the value in $\dfrac{\cos x}{\sin x}\times \cos x$ and get
$\dfrac{{{\cos }^{2}}x}{\sin x}=\dfrac{\cos x}{\sin x}\times \cos x=\cot x\cos x$.
Thus verified $\text{cosec}x-\sin x=\cos x\cot x$.
We verify the result with an arbitrary value of x. Let's assume $x={{60}^{\circ }}$.
Therefore, the result of the left-hand side equation will be $\text{cosec}60-\sin 60$. The value will be
$\text{cosec}60-\sin 60=\dfrac{2}{\sqrt{3}}-\dfrac{\sqrt{3}}{2}=\dfrac{4-3}{2\sqrt{3}}=\dfrac{1}{2\sqrt{3}}$.
The result of the left-hand side equation will be $\cos 60\cot 60$. The value will be
$\cos 60\cot 60=\dfrac{1}{2}\times \dfrac{1}{\sqrt{3}}=\dfrac{1}{2\sqrt{3}}$.
Thus, we verify the result $\text{cosec}x-\sin x=\cos x\cot x$.

Note: The relation can be proved both ways. The most preferred one being L.H.S to R.H.S proving. We need to remember that the relation $1-{{\sin }^{2}}x={{\cos }^{2}}x$ is valid only in their common domain $0\le x\le \dfrac{\pi }{2}$. Rest of the relations are valid for the full domain of $\mathbb{R}$.