Q.4 (1) \(\dfrac{1}{\sin^2\theta} - \dfrac{1}{\cos^2\theta} - \dfrac{1}{\tan^2\theta} - \dfrac{1}{\cot^2\theta} - \dfrac{1}{\sec^2\theta} - \dfrac{1}{\operatorname{cosec}^2\theta} = -3\), then find the value of θ.
Best answer
Answer: θ = 45º
\(\dfrac{1}{\sin^2\theta} = \operatorname{cosec}^2\theta\), \(\dfrac{1}{\cos^2\theta} = \sec^2\theta\), \(\dfrac{1}{\tan^2\theta} = \cot^2\theta\), \(\dfrac{1}{\cot^2\theta} = \tan^2\theta\), \(\dfrac{1}{\sec^2\theta} = \cos^2\theta\), \(\dfrac{1}{\operatorname{cosec}^2\theta} = \sin^2\theta\).
∴ \(\operatorname{cosec}^2\theta - \sec^2\theta - \cot^2\theta - \tan^2\theta - \cos^2\theta - \sin^2\theta = -3\)
∴ \((\operatorname{cosec}^2\theta - \cot^2\theta) - (\sec^2\theta + \tan^2\theta) - (\sin^2\theta + \cos^2\theta) = -3\)
∴ \(1 - (\sec^2\theta + \tan^2\theta) - 1 = -3\) (using \(\operatorname{cosec}^2\theta - \cot^2\theta = 1\) and \(\sin^2\theta + \cos^2\theta = 1\))
∴ \(\sec^2\theta + \tan^2\theta = 3\)
∴ \((1 + \tan^2\theta) + \tan^2\theta = 3\), so \(2\tan^2\theta = 2\) and \(\tan^2\theta = 1\).
As θ is acute, tan θ = 1, giving θ = 45º.
Check: \(2 - 2 - 1 - 1 - \dfrac12 - \dfrac12 = -3\). ✓
Log in or create a free account to answer this question.
