Respuesta :

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[tex]\csc(-x)=\dfrac1{\sin(-x)}=-\dfrac1{\sin x}=-\csc x[/tex]

[tex]1+\tan^2x=\sec^2x[/tex]

So,

[tex]\dfrac{\csc(-x)}{1+\tan^2x}=-\dfrac{\csc x}{\sec^2x}=-\dfrac{\cos^2x}{\sin x}=-\cot x\cos x[/tex]

So the answer is C.

Answer:

The simplified expression is:

               Option: c

           c.     -cos(x)cot(x)

Step-by-step explanation:

We are asked to simplify the expression:

[tex]\dfrac{\csc (-x)}{1+\tan^2 x}[/tex]

We know that :

           [tex]\csc (-x)=-\csc x[/tex]

Also, we know that:

[tex]\sec^2 x-\tan^2 x=1\\\\i.e.\\\\\sec^2 x=1+\tan^2 x[/tex]

i.e.

[tex]\dfrac{\csc (-x)}{1+\tan^2 x}=\dfrac{-\csc x}{\sec^2 x}[/tex]

Also, we know that:

[tex]\csc x=\dfrac{1}{\sin x}[/tex]

and

[tex]\sec x=\dfrac{1}{\cos x}\\\\\\i.e.\\\\\\\sec^2 x=\dfrac{1}{\cos^2 x}\\\\i.e.\\\\\cos^2x=\dfrac{1}{\sec^2 x}[/tex]

Hence, we have:

[tex]\dfrac{\csc (-x)}{1+\tan^2 x}=-\dfrac{\cos^2 x}{\sin x}[/tex]

which could also be written as:

[tex]\dfrac{\csc (-x)}{1+\tan^2 x}=-\dfrac{\cos x}{\sin x}\times \cos x[/tex]

Now, we have:

[tex]\cot x=\dfrac{\cos x}{\sin x}[/tex]

Hence, we get:

[tex]\dfrac{\csc (-x)}{1+\tan^2 x}=-\cos x\cot x[/tex]