why is -csc x cot x= -cos x csc^2x?

Question

i thought deriv of cos was -sin ....

Answer 1

This is not a derivative question. In order to see how these are equal, you need to write the equation in terms of sine and cosine. A couple things you should know:

cot x = cos x / sin x

1 / sin x = sec x

So, let's write the equation in terms of the basic trig functions, sine and cosine:

-csc x cot x = -cos x csc² x
-(1 / sin x)(cos x / sin x) = -cos x (1 / sin² x)
-(cos x / sin² x) = -cos x (1 / sin² x)
-cos x (1 / sin² x) = -cos x (1 / sin² x)
-cos x csc² x = -cos x csc² x

We can see that simply by rewriting the terms in its basic form, we can put both equations in the form that was given on the right side inititally.

Answer 2

-csc x cot x = -cos x csc^2x

-(1/sin x)(cos x/sin x) = -cos x csc^2x

-(1)(cos x/sin² x) = -cos x csc^2x

-cos x (1/sin² x) = -cos x csc^2x

-cos x csc² x = -cos x csc^2x
.

Answer 3

csc(x) = 1/sin(x)
cot(x) = cos(x)/sin(x)

expression on the left is equal to

- cos(x) / [sin(x)]^2 or

-cos(x) * (1/sin(x))*(1/sin(x))

-cos(x) * csc(x)^2

good luck.

Answer 4

LHS
= -cosec x.cot x
= (- 1 / sin x).(cos x / sinx)
= - cos x .(1/ sin²x)
= - cos x.cosec²x

RHS
= - cos x.cosec²x

LHS = RHS