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Question Number 140192 by meetbhavsar25 last updated on 05/May/21

Answered by mr W last updated on 05/May/21

$$say{\tan }^{-1}x=t$$$$x=\tan t$$$$\cos t=\tan t=\frac{\sin t}{\cos t}$$$${\cos }^{2}t=\sin t$$$${\sin }^{2}t+\sin t-1=0$$$$\sin t=\frac{-1+\sqrt{5}}{2}$$$$\frac{x^2}{2}=\frac{{\tan }^{2}t}{2}=\frac{{\sin }^{2}t}{{2\cos }^{2}t}=\frac{\sin t}{2}=\frac{\sqrt{5}-1}{4}$$$$\Rightarrow {\cos }^{-1}\left(\frac{x^2}{2}\right)={\cos }^{-1}\left(\frac{\sqrt{5}-1}{4}\right)=\frac{2\pi }{5}=72°$$

Commented by mr W last updated on 05/May/21

$$\cos \left(\frac{x^2}{2}\right){doesn}^{\prime }tmakemuchsense,$$$$soichangedthequestionto{\cos }^{-1}\left(\frac{x^2}{2}\right).$$

Commented by meetbhavsar25 last updated on 05/May/21

Thank you and you were correct....i mistakenly wrote cos instead of cos-¹

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