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Question-211956




Question Number 211956 by Durganand last updated on 25/Sep/24
Answered by Frix last updated on 25/Sep/24
tan α =t     tan 2α =((2t)/(1−t^2 ))     sin 2α =((2t)/(1+t^2 ))  (1/t)−((1−t^2 )/(2t))=((2−(1−t^2 ))/(2t))=((1+t^2 )/(2t))=(1/(sin 2α))
$$\mathrm{tan}\:\alpha\:={t}\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha} \\ $$

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