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Question Number 144914 by bobhans last updated on 30/Jun/21

$$\mathrm{If}\frac{1+\tan 4\sqrt{\theta }}{1-\tan 4\sqrt{\theta }}=\tan \theta ,\mathrm{then}\mathrm{find}\mathrm{possible}$$$$\mathrm{value}\mathrm{of}\tan (\theta +11\sqrt{\theta }).$$

Answered by imjagoll last updated on 30/Jun/21

Answered by mr W last updated on 30/Jun/21

$$\frac{\tan \frac{\pi }{4}+\tan 4\sqrt{\theta }}{1-\tan \frac{\pi }{4}\times \tan 4\sqrt{\theta }}=\tan \theta$$$$\tan (\frac{\pi }{4}+4\sqrt{\theta })=\tan \theta$$$$k\pi +\frac{\pi }{4}+4\sqrt{\theta }=\theta$$$$k\pi +\frac{\pi }{4}+4={(\sqrt{\theta }-2)}^2$$$$\sqrt{\theta }=2+\sqrt{k\pi +\frac{\pi }{4}+4}$$$$\Rightarrow k⩾-1$$$$\sqrt{\theta }=2-\sqrt{k\pi +\frac{\pi }{4}+4}$$$$\Rightarrow k=-1$$$$$$$$\theta +11\sqrt{\theta }=k\pi +\frac{\pi }{4}+15\sqrt{\theta }$$$$\theta +11\sqrt{\theta }=k\pi +\frac{\pi }{4}+30\pm 15\sqrt{k\pi +\frac{\pi }{4}+4}$$$$\Rightarrow \tan (\theta +11\sqrt{\theta })=\tan (\frac{\pi }{4}+30+15\sqrt{k\pi +\frac{\pi }{4}+4})withk⩾-1$$$$or$$$$\Rightarrow \tan (\theta +11\sqrt{\theta })=\tan (\frac{\pi }{4}+30-15\sqrt{-\pi +\frac{\pi }{4}+4})$$

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