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

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

Explanation

Solution

The given trigonometric is csc⁡(−x)sec⁡(−x)\dfrac{{\csc ( - x)}}{{\sec ( - x)}}
An even function is symmetric (by reflection) about the yy-axis, i.e. f⁡(−x)=f(x)\operatorname{f} ( - x) = f(x)
An odd function is symmetric (by 180∘{180^ \circ } rotation) about the origin, i.e.f⁡(−x)=−f(x)\operatorname{f} ( - x) = - f(x)
Use the even and odd properties trigonometric functions.
sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x And cos⁡(−x)=cos⁡x\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 csc⁡(−x)sec⁡(−x)\dfrac{{\csc ( - x)}}{{\sec ( - x)}}
We verify that the identity is csc⁡(−x)sec⁡(−x)=−cot⁡x\dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x
Let’s take the LHS (Left Hand Side)
⇒csc⁡(−x)sec⁡(−x)\Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}}
Use the even and odd properties of trigonometric functions, hence we get
sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x And cos⁡(−x)=cos⁡x\cos ( - x) = \cos x
csc⁡(−x)\csc ( - x)Andsec⁡(−x)\sec ( - x) the formula is,
⇒csc⁡(−x)=1sin⁡(−x)\Rightarrow \csc ( - x) = \dfrac{1}{{\sin ( - x)}} And
⇒sec⁡(−x)=1cos⁡(−x)\Rightarrow \sec ( - x) = \dfrac{1}{{\cos ( - x)}}
Now the two formula substitute in thecsc⁡(−x)sec⁡(−x)\dfrac{{\csc ( - x)}}{{\sec ( - x)}}, hence we get
⇒csc⁡(−x)sec⁡(−x)=1sin⁡(−x)1cos⁡(−x)\Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}} = \dfrac{{\dfrac{1}{{\sin ( - x)}}}}{{\dfrac{1}{{\cos ( - x)}}}}
Then the division we rewrite in the form of abcd=ab×dc\dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{a}{b} \times \dfrac{d}{c}, hence we get
⇒1sin⁡(−x)×cos⁡(−x)1\Rightarrow \dfrac{1}{{\sin ( - x)}} \times \dfrac{{\cos ( - x)}}{1}
Hence we use the even and odd properties for trigonometric functions, hence we get
⇒1−sin⁡x×cos⁡x1\Rightarrow \dfrac{1}{{ - \sin x}} \times \dfrac{{\cos x}}{1}
We rewrite the form, hence we get
⇒cos⁡x−sin⁡x\Rightarrow \dfrac{{\cos x}}{{ - \sin x}}
We use the formulacos⁡xsin⁡x=cot⁡x\dfrac{{\cos x}}{{\sin x}} = \cot x, hence we substitute in the function, hence we get
⇒−cot⁡x\Rightarrow - \cot x
⇒csc⁡(−x)sec⁡(−x)=−cot⁡x\Rightarrow \dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x
Hence verify that the identity csc⁡(−x)sec⁡(−x)=−cot⁡x\dfrac{{\csc ( - x)}}{{\sec ( - x)}} = - \cot x.

Note: An even function is symmetric (by reflection) about the yy-axis, i.e.f⁡(−x)=f(x)\operatorname{f} ( - x) = f(x)
An odd function is symmetric (by180∘{180^ \circ } rotation) about the origin, i.e.f⁡(−x)=−f(x)\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)\cos ( - x) = \cos (x)
The Secant function is even sec⁡(−x)=sec⁡(x)\sec ( - x) = \sec (x)
Odd trigonometric functions and identities:
The Sine function is odd sin⁡(−x)=−sin⁡(x)\sin ( - x) = - \sin (x)
The Cosecant function is odd csc⁡(−x)=−csc⁡(x)\csc ( - x) = - \csc (x)
The Tangent function is odd tan⁡(−x)=−tan⁡(x)\tan ( - x) = - \tan (x)
The Cotangent function is odd cot⁡(−x)=−cot⁡(x)\cot ( - x) = - \cot (x)