If mtan⁡(α−θ)cos⁡2θ=ntan⁡θcos⁡2(α−θ)\frac{m \tan(\alpha - \theta)}{\cos^2 \theta} = \frac{n \tan \theta}{\cos^2(\alpha - \theta)}…

Problem

If mtan⁡(α−θ)cos⁡2θ=ntan⁡θcos⁡2(α−θ)\frac{m \tan(\alpha - \theta)}{\cos^2 \theta} = \frac{n \tan \theta}{\cos^2(\alpha - \theta)} Prove that : θ=12[α−tan⁡−1(n−mn+mtan⁡α)]\theta = \frac{1}{2} \left[ \alpha - \tan^{-1} \left( \frac{n-m}{n+m} \tan \alpha \right) \right]

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