Verify each identity:
1. (sin α-csc α)² = cot²α-cos²α
2. 1-tan²A÷1+tan²A=2cos²A-1
3. (cos α - sec α ÷ sec α)+(sin α - csc α ÷ csc α) = -1

1.

(sin α- csc α)²

= (sin α - 1/sin α)²

= sin²α - 2 (sin α)( 1/sin α) + 1/sin²α

= sin²α - 2  + 1/sin²α

= (sin²α - 1)  + (1/sin²α - 1)

= -(1 - sin²α ) + (1 - sin²α)/sin²α

= - cos²α  + cos²/sin²α

= cot²α-cos²α

2.

(1-tan²A)/(1+tan²A)

= (1 - sin²A / cos²A ) / (1 + sin²A/cos²A)

= (cos²A - sin²A)/(cos²A + sin²A)

= (cos²A - (1 - cos²A))/(1)

= 2 cos²A - 1

3.

(cos α - sec α)/sec α + (sin α - csc α)/csc α

= (cos α - sec α)/sec α + (sin α - csc α) /csc α

= cos α/sec α - 1  +sin α /csc α - 1

= cos  α / (1 / cos  α) + sin α /(1 / sin α) - 2

= cos²α + sin²α - 2

= 1 - 2

= -1

answer: the two figures are congruent.

step-by-step explanation: they're congruent because figure efgh is reflected across the x and y axis but it was also rotated.

hope it !