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If $\cos^2 A=\dfrac{a^2-1}{3}$ and $\tan^2\left(\dfrac{A}{2}\right)=\tan^{2/3} B$. Then find $\cos^{2/3}B+\sin^{2/3}B $.


I tried componendo and dividendo to write the second statement as cos A but i couldnt simplify it

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  • $\begingroup$ Write your question properly. @Himanshu Arora $\endgroup$ – Shodharthi Sep 18 '14 at 8:53
  • $\begingroup$ Your question is difficult to read. Did you mean if $\cos^2 A = \dfrac{a^2 - 1}{3}$ and $\tan^2\left(\dfrac{A}{2}\right) = \tan^{\frac{2}{3}} B$ find $\cos^{\frac{2}{3}} B + \sin^{\frac{2}{3}} B$? $\endgroup$ – N. F. Taussig Sep 18 '14 at 9:38
  • $\begingroup$ yes that's the question i want to ask $\endgroup$ – Himanshu Arora Sep 18 '14 at 9:51
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$$\cos^{2/3}B+\sin^{2/3}B=\cos^{2/3}B\left(1+\tan^{2/3}B\right)=\cos^{2/3}B\left(1+\tan^2\frac{A}{2}\right)=\left(\cos^2B\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(\frac{1}{1+\tan^6\left(\frac{A}{2}\right)}\right)^{1/3}\left(1+\tan^2\frac{A}{2}\right)=\left(1+\frac{3\tan^2 \left(\frac{A}{2}\right)\left(1+\tan^2\left(\frac{A}{2}\right)\right)}{1+\tan^6\left(\frac{A}{2}\right)}\right)^{1/3}$$ I hope you can continue from here (you just need to find $\tan^2\frac{A}{2}$ as a function of $a$).

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