Asked in: March 2024 board paperIf \(\dfrac{1}{\sin^2\theta} - \dfrac{1}{\cos^2\theta} - \dfrac{1}{\tan^2\theta} - \dfrac{1}{\cot^2\theta} - \dfrac{1}{\sec^2\theta} - \dfrac{1}{\operatorname{cosec}^2\theta} = -3\), then find the value of θ.

Answer: θ = 45°

Write each term with its reciprocal ratio:

\(\operatorname{cosec}^2\theta - \sec^2\theta - \cot^2\theta - \tan^2\theta - \cos^2\theta - \sin^2\theta = -3\)

Group the terms: \((\operatorname{cosec}^2\theta - \cot^2\theta) - (\sec^2\theta + \tan^2\theta) - (\sin^2\theta + \cos^2\theta) = -3\)

\(\therefore 1 - (\sec^2\theta + \tan^2\theta) - 1 = -3\) \(\therefore \sec^2\theta + \tan^2\theta = 3\)

\(\therefore (1 + \tan^2\theta) + \tan^2\theta = 3\) \(\therefore 2\tan^2\theta = 2\) \(\therefore \tan^2\theta = 1\)

\(\therefore \tan\theta = 1\) (θ acute) \(\therefore \theta = 45^\circ\).