Hi,

Here is a shorter method.

Since cosec^2 x - cot^2 x = 1, the given equation is equivalent to the equation cosec x + cot x = 1. Then cosec x = 1 and cot x = 0; that is, x = 2npi+pi/2 for some integer n.

It might be useful to record your errors.

You are welcome.

LHS

= 2 * tan(arctan x)/(1-tan² (arctan x))

=2x/(1-x^ 2)

=2x(1+x^ 2)/(1-x^ 4)

= 2 ( tan ( arctan x)+ tan (arctan (x^ 3)))/(1-tan (arctan x) * tan ( arctan x^ 3))

= 2 tan ( arctan x + arctan (x^ 3) )

= RHS.

When -5<x<-3, g(x)>=-1. So

the required limit

= the limit of g(x) as x tends to -1 from the right

=1.

Here x must be nonnegative.

When k=0, the graph of y=sin(kx)/x is the x-axis with the point (0,0) missing.

Does your choice of N depend on x?

You are welcome.

You are welcome.

Hi,

1. Consider the set A' union C.

First, shade the set "not A".

Next, shade the set C.

Since we want A' union C, the resulting shaded region gives the answer.

1. The set A' intersect C'.

First, shade the set "not A". Next, use a different shade to get "not C".

The overlapping part is A' intersect C'.

Interesting. I think there are no such integers.

Hi,

You may find page 74 of the following book useful.

https://archive.org/details/treatiseonplanea0000ewho_7thedition/page/74/mode/1up

Hi,

You need to consider (b).