Question

How do you verify the identity $\dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x$?

Answer 1

Hint: The given trigonometric is $\dfrac{{\csc ( - x)}}{{\sec ( - x)}}$
An even function is symmetric (by reflection) about the $y$-axis, i.e. $\operatorname{f} ( - x) = f(x)$
An odd function is symmetric (by ${180^ \circ }$ rotation) about the origin, i.e.$\operatorname{f} ( - x) = - f(x)$
Use the even and odd properties trigonometric functions.
$\sin ( - x) = - \sin x$ And $\cos ( - x) = \cos x$
We use even and odd properties of trigonometric functions after that substitution.
After that we simplify the trigonometric function.
Finally we get the proof of identities in the given trigonometric function.

Complete step-by-step solution:
The given trigonometric is $\dfrac{{\csc ( - x)}}{{\sec ( - x)}}$
We verify that the identity is $\dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x$
Let’s take the LHS (Left Hand Side)
$ \Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}}$
Use the even and odd properties of trigonometric functions, hence we get
$\sin ( - x) = - \sin x$ And $\cos ( - x) = \cos x$
$\csc ( - x)$And$\sec ( - x)$ the formula is,
$ \Rightarrow \csc ( - x) = \dfrac{1}{{\sin ( - x)}}$ And
$ \Rightarrow \sec ( - x) = \dfrac{1}{{\cos ( - x)}}$
Now the two formula substitute in the$\dfrac{{\csc ( - x)}}{{\sec ( - x)}}$, hence we get
$ \Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}} = \dfrac{{\dfrac{1}{{\sin ( - x)}}}}{{\dfrac{1}{{\cos ( - x)}}}}$
Then the division we rewrite in the form of $\dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{a}{b} \times \dfrac{d}{c}$, hence we get
$ \Rightarrow \dfrac{1}{{\sin ( - x)}} \times \dfrac{{\cos ( - x)}}{1}$
Hence we use the even and odd properties for trigonometric functions, hence we get
$ \Rightarrow \dfrac{1}{{ - \sin x}} \times \dfrac{{\cos x}}{1}$
We rewrite the form, hence we get
$ \Rightarrow \dfrac{{\cos x}}{{ - \sin x}}$
We use the formula$\dfrac{{\cos x}}{{\sin x}} = \cot x$, hence we substitute in the function, hence we get
$ \Rightarrow - \cot x$
$ \Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x$
Hence verify that the identity $\dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x$.

Note: An even function is symmetric (by reflection) about the $y$-axis, i.e.$\operatorname{f} ( - x) = f(x)$
An odd function is symmetric (by${180^ \circ }$ rotation) about the origin, i.e.$\operatorname{f} ( - x) = - f(x)$
The following shows the even trigonometric functions and odd trigonometric functions.
Even trigonometric functions and identities:
The Cosine function is even $\cos ( - x) = \cos (x)$
The Secant function is even $\sec ( - x) = \sec (x)$
Odd trigonometric functions and identities:
The Sine function is odd $\sin ( - x) = - \sin (x)$
The Cosecant function is odd $\csc ( - x) = - \csc (x)$
The Tangent function is odd $\tan ( - x) = - \tan (x)$
The Cotangent function is odd $\cot ( - x) = - \cot (x)$