If \(∠P + ∠Q = 90^\circ\), then prove that \[ \sqrt{\frac{\sin P}{\cos Q}} - \sin P \cos Q = \cos^2 P \]

Q.If \(∠P + ∠Q = 90^\circ\), then prove that \[ \sqrt{\frac{\sin P}{\cos Q}} - \sin P \cos Q = \cos^2 P \]

\[ \sqrt{\frac{\sin P}{\cos Q}} - \sin P \cos Q \] \[ = \sqrt{\frac{\sin P}{\cos(90^\circ - P)}} - \sin P \cos(90^\circ - P) \quad [\because P + Q = 90^\circ \Rightarrow Q = 90^\circ - P] \] \[ = \sqrt{\frac{\sin P}{\sin P}} - \sin P \cdot \sin P = 1 - \sin^2 P = \cos^2 P \quad \text{(Proved)} \]