Question Number 140192 by meetbhavsar25 last updated on 05/May/21
Answered by mr W last updated on 05/May/21
$${say}\:\mathrm{tan}^{−\mathrm{1}} {x}={t} \\ $$$${x}=\mathrm{tan}\:{t} \\ $$$$\mathrm{cos}\:{t}=\mathrm{tan}\:{t}=\frac{\mathrm{sin}\:{t}}{\mathrm{cos}\:{t}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{t}=\mathrm{sin}\:{t} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{t}+\mathrm{sin}\:{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{tan}^{\mathrm{2}} \:{t}}{\mathrm{2}}=\frac{\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{2cos}^{\mathrm{2}} \:{t}}=\frac{\mathrm{sin}\:{t}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{2}\pi}{\mathrm{5}}=\mathrm{72}° \\ $$
Commented by mr W last updated on 05/May/21
$$\mathrm{cos}\:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:{doesn}'{t}\:{make}\:{much}\:{sense}, \\ $$$${so}\:{i}\:{changed}\:{the}\:{question}\:{to}\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right). \\ $$
Commented by meetbhavsar25 last updated on 05/May/21
Thank you and you were correct....i mistakenly wrote cos instead of cos-¹