Verified answer

From the triangle, a = 5, b = 12 and c = 13.

sinA = 5/13

cosA = 12/13

LHS = cot²A

= cos²A/sin²A

= (12/13)² / (5/13)²

= 12²/5²

= 144/25

RHS = csc²A - 1

= (1/sin²A) - 1

= 1/{(5/13)²} - 1

= (13²/5²) - 1

= 169/25 - 1

= 169/25 - 25/25

= 144/25

Hence, the equation cot²A = csc²A - 1 is true.

Draw the figure.

In the right triangle ABC,

a = 5, b = 12 and hypotenuse = c = 13

5² + 12² = 13²

=> a² + b² = c²

b² = c² - a²

=> b²/a² = c²/a² - a²/a²

=> (b/a)² = (c/a)² - 1

=> (cot A)² = (csc A)² - 1

=> cot² A = csc² A - 1

Numerically, LHS = (12/5)²

RHS = (13/5)² - 1 = (13² - 5²)/5²

= 12²/5² = (12/5)²

Hence LHS = RHS

Cot^2 a=(12/5)^2=144/25=(169-25)/25=169/25-1=(13/5)^2-1=csc^2-1

cot² A = csc² A – 1

NOTE sin²A + cos²A = 1 then cos²A=1-sin²A

csc² A - 1 = 1/(sin²A) - 1 = (1-sin²A)/sin²A = cos²A/sin²A = cot²A

dnadan1

sin^2(A) + cos^2(A) = 1... divide all terms by sin(A)

==> 1 + cot^2(A) = csc^2(A)

cot^2(A) = cos^2(A)/sin^2(A)

= [1-sin^2(A)]/sin^2(A)

= csc^2(A) - 1