Here is the answer to your question.
Given: A circle with centre O. PQ and PR are tangents drawn from an external point P to the circle.
To prove: ∠QPR = 2∠OQR
∠OQP = ∠ORP = 90° (Radius is perpendicular to the tangent at point of contact)
In quadrilateral OQPR
∠QPR + ∠OQP + ∠QOR + ∠ORP = 360°
∴∠OPR + 90° + ∠QOR + 90° = 360°
⇒ ∠OPR + ∠QOR = 180° ... (1)
OQ = OR (Radius of the circle)
⇒ ∠ORQ = ∠OQR (Equal sides have equal angles opposite to them)
∠QOR + ∠OQR + ∠ORQ = 180° (angle sum property)
⇒ ∠QOR + 2∠OQR = 180° ... (2)
From (1) and (2), we get
∠QPR + ∠QOR = ∠QOR + 2∠OQR
⇒ ∠QPR = 2∠OQR