Q.Given: \(\tan \theta + \sin \theta = m\) and \(\tan \theta - \sin \theta = n\) Prove that: \[ m^2 - n^2 = 4\sqrt{mn} \quad \text{where } 0^\circ < \theta < 90^\circ \]

Left-hand side = \(m^2 - n^2\) \[ = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 = 4 \tan \theta \sin \theta \] Right-hand side = \(4\sqrt{mn}\) \[ = 4\sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)} = 4\sqrt{\tan^2 \theta - \sin^2 \theta} = 4\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta} = 4\sqrt{\sin^2 \theta \left(\frac{1}{\cos^2 \theta} - 1\right)} = 4\sqrt{\sin^2 \theta (\sec^2 \theta - 1)} = 4\sqrt{\sin^2 \theta \tan^2 \theta} = 4 \tan \theta \sin \theta \] Therefore, Left-hand side = Right-hand side (Proved)
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