Sun^4A+sin^2Acos^2A=sin^2A

Step-by-step explanation:

sin⁴A + sin²Acos²A = sin²A

Proving the left hand side ( LHS)

That's

sin⁴A can be written as ( sin²A)(sin ²A)

So we have

( sin²A)(sin ²A) + sin²Acos²A

Next factor sin²A out

That's

sin²A ( sin²A + cos²A)

Using trigonometric identities

That's

Simplify the expression

That's

sin²A × 1

We have the final answer as

As proven

Hope this helps you

b. ol

step-by-step explanation:

i got the answer right on edg.

congruent: (geometry) (of figures) identical in form; coinciding exactly when superimposed.

place r′ so that qr′ is parallel to the tangent.

now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.

also, because of the symmetry, ∠r′py=∠qpx.

finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.

stringing all these equalities together, we end up with ∠prq=∠qpx.

this picture will show how its going to look like.

hope this by jack monhen