Question
Prove that:$$2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \left(\frac{\theta}{2}\right)\right)=\cos ^{-1}\left(\frac{b+a \cos \theta}{a+b \cos \theta}\right)$$
Step 1
So, we have: $$ 2 \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \left(\frac{\theta}{2}\right)\right)=\cos ^{-1}\left(\frac{1-\left(\sqrt{\frac{a-b}{a+b}} \tan \left(\frac{\theta}{2}\right)\right)^2}{1+\left(\sqrt{\frac{a-b}{a+b}} \tan Show more…
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