If you have a two digit divisor in a three digit dividend is the quotient always have the same number of digits

x = π/2 + πk

Step-by-step explanation:

cot² x csc² x + 2 csc² x − cot² x = 2

Multiply both sides by sin² x:

cot² x + 2 − cos² x = 2 sin² x

Add cos² x to both sides:

cot² x + 2 = 2 sin² x + cos² x

Pythagorean identity:

cot² x + 2 = sin² x + 1

Subtract 1 from both sides:

cot² x + 1 = sin² x

Pythagorean identity:

csc² x = sin² x

Multiply both sides by sin² x:

1 = sin⁴ x

Take the fourth root:

sin x = ±1

Solve for x:

x = π/2 + 2πk, 3π/2 + 2πk

Which simplifies to:

x = π/2 + πk