Find the general solution of \(\tan^2\theta = 1\).
Q.5 Find the general solution of \(\tan^2\theta = 1\).
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Answer: \(\theta = n\pi \pm \dfrac{\pi}{4}\), \(n \in Z\).
\(\tan^2\theta = 1 = \tan^2\dfrac{\pi}{4}\).
The general solution of \(\tan^2\theta = \tan^2\alpha\) is \(\theta = n\pi \pm \alpha\). Hence \(\theta = n\pi \pm \dfrac{\pi}{4}\), where \(n \in Z\).
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· 28 Oct 2026
AI-verified · Oct 2026
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