Question

Simplify the expression. csc(-x)/1+tan^2(x)

Answer 1

Answer:

The simplified expression is:

               Option: c

           c.     -cos(x)cot(x)

Step-by-step explanation:

We are asked to simplify the expression:

$\dfrac{\csc (-x)}{1+\tan^2 x}$

We know that :

           $\csc (-x)=-\csc x$

Also, we know that:

$\sec^2 x-\tan^2 x=1\\\\i.e.\\\\\sec^2 x=1+\tan^2 x$

i.e.

$\dfrac{\csc (-x)}{1+\tan^2 x}=\dfrac{-\csc x}{\sec^2 x}$

Also, we know that:

$\csc x=\dfrac{1}{\sin x}$

and

$\sec x=\dfrac{1}{\cos x}\\\\\\i.e.\\\\\\\sec^2 x=\dfrac{1}{\cos^2 x}\\\\i.e.\\\\\cos^2x=\dfrac{1}{\sec^2 x}$

Hence, we have:

$\dfrac{\csc (-x)}{1+\tan^2 x}=-\dfrac{\cos^2 x}{\sin x}$

which could also be written as:

$\dfrac{\csc (-x)}{1+\tan^2 x}=-\dfrac{\cos x}{\sin x}\times \cos x$

Now, we have:

$\cot x=\dfrac{\cos x}{\sin x}$

Hence, we get:

$\dfrac{\csc (-x)}{1+\tan^2 x}=-\cos x\cot x$

Answer 2

$\csc(-x)=\dfrac1{\sin(-x)}=-\dfrac1{\sin x}=-\csc x$

$1+\tan^2x=\sec^2x$

So,

$\dfrac{\csc(-x)}{1+\tan^2x}=-\dfrac{\csc x}{\sec^2x}=-\dfrac{\cos^2x}{\sin x}=-\cot x\cos x$

So the answer is C.