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Question Number 186125 by Mingma last updated on 01/Feb/23

Answered by som(math1967) last updated on 01/Feb/23

$$let{2}^{sin\alpha }=3^{sin\beta }=5^{sin\theta }=k$$$$\Rightarrow k^{\frac{1}{sin\alpha }}=2\Rightarrow k^{cosec\alpha }=2$$$$\Rightarrow k^{cosec\beta }=3$$$$\Rightarrow k^{\mathrm{cosec}\theta }=5$$$$k^{\mathrm{cosec}\alpha +\mathrm{cosec}\beta +\mathrm{cosec}\theta }=2\times 3\times 5$$$$\Rightarrow k^8=30\therefore k={\left(30\right)}^{\frac{1}{8}}$$$$2^{sin\alpha }+2^{\sin \beta }+2^{\sin \theta }$$$$=k+k+k$$$$=3k=3\times {\left(30\right)}^{\frac{1}{8}}$$

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