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