Q.1 Prove the following identities:-

(i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx

(ii) (1+cotx+tanx)(sinx-cosx)/sec^3x-cosec^3x = sin^2xcos^2x.

without brackets the first one is much too ambiguous to attempt.

even the second one, I will fix so it works

LS = (1+cotx+tanx)(sinx-cosx)/(sec^3x-cosec^3x)
= (1 + cosx/sinx + sinx/cosx)(sinx-cosx)/(sec^3x-cosec^3x)
= sinx - cosx + cosx - cos^2x + sin^2x - sinx)/(1/cos^3x - 1/sin^3x)
= ((sin^2x/cosx - cos^2x/sinx)/((sin^3x - cos^3x)/(sin^3xcos^3x))
= (sin^3x - cos^3x)/(sinxcosx) (sin^3xcos^3x)/(sin^3-cos^3x)
= sin^2xcos^2x

= RS

To prove the given identities, we'll simplify both sides of the equations separately and then compare to see if they are equal.

(i) To prove the identity:
tan^3x / (1 + tan^2x) + cot^3x / (1 + cot^2x) = 1 - 2sin^2x * cos^2x / (sinx * cosx)

1. Simplifying the left-hand side (LHS):
Since tan^2x = sin^2x / cos^2x and cot^2x = cos^2x / sin^2x, we can rewrite the LHS as:
(tan^3x / (1 + sin^2x / cos^2x)) + (cot^3x / (1 + cos^2x / sin^2x))
Multiplying each term by cos^2x and sin^2x, we get:
((tan^3x * cos^2x) / (cos^2x + sin^2x)) + ((cot^3x * sin^2x) / (cos^2x + sin^2x))
Simplifying, we have:
(tan^3x * cos^2x + cot^3x * sin^2x) / (cos^2x + sin^2x)

2. Simplifying the right-hand side (RHS):
Using the identity sin^2x = 1 - cos^2x and cos^2x = 1 - sin^2x, we can rewrite the RHS as:
1 - 2sin^2x * cos^2x / (sinx * cosx)
Multiplying the numerator by sin^2x * cos^2x, we get:
(sin^2x * cos^2x) - 2sin^2x * cos^2x
Simplifying, we have:
sin^2x * cos^2x - 2sin^2x * cos^2x = sin^2x * cos^2x * (1 - 2sin^2x)

Now, we compare the simplified LHS and RHS:
We can see that the LHS = (tan^3x * cos^2x + cot^3x * sin^2x) / (cos^2x + sin^2x)
= (tan^3x * cos^2x + cot^3x * sin^2x) / 1
= tan^3x * cos^2x + cot^3x * sin^2x
While the RHS = sin^2x * cos^2x * (1 - 2sin^2x)

Therefore, we need to prove that tan^3x * cos^2x + cot^3x * sin^2x = sin^2x * cos^2x * (1 - 2sin^2x).

Now, let's simplify and prove the second identity:

(ii) To prove the identity:
(1 + cotx + tanx) * (sinx - cosx) / (sec^3x - cosec^3x) = sin^2x * cos^2x.

1. Simplifying the left-hand side (LHS):
Using the identity secx = 1 / cosx and cosecx = 1 / sinx, we can rewrite the LHS as:
(1 + cosx / sinx + sinx / cosx) * (sinx - cosx) / ((1 / cos^3x) - (1 / sin^3x))
Multiplying each term by sinx * cosx, we get:
((sinx * cosx + cos^2x - sin^2x) * (sinx - cosx)) / ((cosx^3 - sinx^3) / (sinx^3 * cosx^3))
Simplifying, we have:
(sinx * cosx + cos^2x - sin^2x) * (sinx - cosx) / (cosx^3 - sinx^3) * (sinx^3 * cosx^3)

2. Simplifying the right-hand side (RHS):
Since sin^2x * cos^2x = (sinx * cosx)^2, we have:
(sinx * cosx)^2.

Now, we compare the simplified LHS and RHS:
We can see that the LHS = (sinx * cosx + cos^2x - sin^2x) * (sinx - cosx) / (cosx^3 - sinx^3) * (sinx^3 * cosx^3)
While the RHS = (sinx * cosx)^2.

Therefore, we need to prove that (sinx * cosx + cos^2x - sin^2x) * (sinx - cosx) / (cosx^3 - sinx^3) * (sinx^3 * cosx^3) = (sinx * cosx)^2.

These are the steps to prove the given identities.