Find the value of \(\sqrt{\frac{\sec^2\theta + \mathrm{cosec}^2\theta}{\tan^2\theta - \sin^2\theta}}\).
cosec3 θ
Numerator: \(\sec^2\theta + \mathrm{cosec}^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}\).
Denominator: \(\tan^2\theta - \sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta = \frac{\sin^2\theta(1 - \cos^2\theta)}{\cos^2\theta} = \frac{\sin^4\theta}{\cos^2\theta}\).
So the fraction = \(\frac{1}{\sin^2\theta\cos^2\theta} \times \frac{\cos^2\theta}{\sin^4\theta} = \frac{1}{\sin^6\theta} = \mathrm{cosec}^6\theta\).
Taking the square root, the value = \(\sqrt{\mathrm{cosec}^6\theta} = \mathrm{cosec}^3\theta\).
Check: at θ = 45°, sec²θ + cosec²θ = 2 + 2 = 4 and tan²θ – sin²θ = 1 – 1/2 = 1/2, so the value = √8 = 2√2; and cosec³ 45° = (√2)3 = 2√2. The values agree.
Hence, cosec3 θ.
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