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Question Number 140368 by meetbhavsar25 last updated on 06/May/21

Answered by benjo_mathlover last updated on 07/May/21

$${\cos }^{-1}\left(\frac{3+5\cos \mathrm{x}}{5+3\cos \mathrm{x}}\right)=\mathrm{u}$$$$\Rightarrow \frac{3+5\cos \mathrm{x}}{5+3\cos \mathrm{x}}=\cos \mathrm{u}$$$$\Rightarrow 3+5\cos \mathrm{x}=5\cos \mathrm{u}+3\cos \mathrm{u}\cos \mathrm{x}$$$$\Rightarrow (5-3\cos \mathrm{u})\cos \mathrm{x}=5\cos \mathrm{u}-3$$$$\Rightarrow \cos \mathrm{x}=\frac{5\cos \mathrm{u}-3}{5-3\cos \mathrm{u}}$$$$\Rightarrow 2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5\cos \mathrm{u}-3}{5-3\cos \mathrm{u}}+\frac{5-3\cos \mathrm{u}}{5-3\cos \mathrm{u}}$$$$\Rightarrow 2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{2\cos \mathrm{u}+2}{5-3\cos \mathrm{u}}$$$$\Rightarrow \frac{1}{2}\sec {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{2\cos \mathrm{u}+2}$$$$\Rightarrow \sec {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{\cos \mathrm{u}+1}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{\cos \mathrm{u}+1}-\frac{\cos \mathrm{u}+1}{\cos \mathrm{u}+1}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{4-4\cos \mathrm{u}}{1+\cos \mathrm{u}}=\frac{4(1-(1-2\sin {}^2\left(\frac{\mathrm{u}}{2}\right))}{2\cos {}^2\left(\frac{\mathrm{u}}{2}\right)}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=4\tan {}^2\left(\frac{\mathrm{u}}{2}\right)$$$$\Rightarrow \tan \left(\frac{\mathrm{x}}{2}\right)=2\tan \left(\frac{\mathrm{u}}{2}\right)$$$$\Rightarrow \frac{\mathrm{u}}{2}={\tan }^{-1}\left(\frac{1}{2}\tan \frac{\mathrm{x}}{2}\right)$$$$\Rightarrow \mathrm{u}={2\tan }^{-1}\left(\frac{1}{2}\tan \frac{\mathrm{x}}{2}\right)$$

Commented by meetbhavsar25 last updated on 07/May/21

$${\cos }^{-1}\left(\frac{3+5\cos \mathrm{x}}{5+3\cos \mathrm{x}}\right)=\mathrm{u}$$$$\Rightarrow \frac{3+5\cos \mathrm{x}}{5+3\cos \mathrm{x}}=\cos \mathrm{u}$$$$\Rightarrow 3+5\cos \mathrm{x}=5\cos \mathrm{u}+3\cos \mathrm{u}\cos \mathrm{x}$$$$\Rightarrow (5-3\cos \mathrm{u})\cos \mathrm{x}=5\cos \mathrm{u}-3$$$$\Rightarrow \cos \mathrm{x}=\frac{5\cos \mathrm{u}-3}{5-3\cos \mathrm{u}}$$$$\Rightarrow 2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5\cos \mathrm{u}-3}{5-3\cos \mathrm{u}}+\frac{5-3\cos \mathrm{u}}{5-3\cos \mathrm{u}}$$$$\Rightarrow 2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{2\cos \mathrm{u}+2}{5-3\cos \mathrm{u}}$$$$Thankyou...muchappreciated$$$$\Rightarrow \frac{1}{2}\sec {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{2\cos \mathrm{u}+2}$$$$\Rightarrow \sec {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{\cos \mathrm{u}+1}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5-3\cos \mathrm{u}}{\cos \mathrm{u}+1}-\frac{\cos \mathrm{u}+1}{\cos \mathrm{u}+1}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{4-4\cos \mathrm{u}}{1+\cos \mathrm{u}}=\frac{4(1-(1-2\sin {}^2\left(\frac{\mathrm{u}}{2}\right))}{2\cos {}^2\left(\frac{\mathrm{u}}{2}\right)}$$$$\Rightarrow \tan {}^2\left(\frac{\mathrm{x}}{2}\right)=4\tan {}^2\left(\frac{\mathrm{u}}{2}\right)$$$$\Rightarrow \tan \left(\frac{\mathrm{x}}{2}\right)=2\tan \left(\frac{\mathrm{u}}{2}\right)$$$$\Rightarrow \frac{\mathrm{u}}{2}={\tan }^{-1}\left(\frac{1}{2}\tan \frac{\mathrm{x}}{2}\right)$$$$\Rightarrow \mathrm{u}={2\tan }^{-1}\left(\frac{1}{2}\tan \frac{\mathrm{x}}{2}\right)$$$$Thankyou$$

Commented by meetbhavsar25 last updated on 07/May/21

$$Ifoundasimplermethod....$$$${\cos }^{-1}\left\{\frac{3+5\cos x}{5+3\cos x}\right\}=y$$$$\frac{\cos y}{1}=\frac{3+5\cos x}{5+3\cos x}$$$$\frac{\cos y+1}{\cos y-1}=\frac{8+8\cos x}{2\cos x-2}(usingcomponendo-dividendo)$$$$\frac{2\cos {}^2\frac{y}{2}}{-2\sin {}^2\frac{y}{2}}=\frac{8\times 2\cos {}^2\frac{x}{2}}{-2\times 2\sin {}^2\frac{x}{2}}$$$$\cot {}^2\frac{y}{2}=4\times \cot {}^2\frac{x}{2}$$$$\tan {}^2\frac{x}{2}=4\times \tan {}^2\frac{y}{2}$$$$\tan \frac{x}{2}=2\times \tan \frac{y}{2}$$$$\tan \frac{y}{2}=\frac{1}{2}\tan \frac{x}{2}$$$$\frac{y}{2}={\tan }^{-1}\left\{\frac{1}{2}\tan \frac{x}{2}\right\}$$$$y={2\tan }^{-1}\left\{\frac{1}{2}\tan \frac{x}{2}\right\}$$

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