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Question Number 51364 by rahul 19 last updated on 26/Dec/18

$$Thevalueofxforwhich$$$$\sin \left({\cot }^{-1}(1+x)\right)=\cos \left({\tan }^{-1}x\right)is?$$

Commented by rahul 19 last updated on 26/Dec/18

$$sin\left({\cot }^{-1}(1+x)\right)=\sin (\frac{\pi }{2}-{\tan }^{-1}x)$$$$\Rightarrow {\tan }^{-1}\left(\frac{1}{1+x}\right)=(\frac{\pi }{2}-{\tan }^{-1}x)$$$$\Rightarrow {tan}^{-1}\left(\frac{1}{1+x}\right)+{\tan }^{-1}x=\frac{\pi }{2}$$$$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{1+x}+x}{1-\frac{x}{1+x}}\right)=\frac{\pi }{2}$$$$\Rightarrow {\tan }^{-1}\left(\frac{x^2+x+1}{1}\right)=\frac{\pi }{2}$$$$.....?????$$

Commented by prakash jain last updated on 26/Dec/18

$${\cot }^{-1}(1+x)=\theta$$$$\cot \theta =1+x$$$$\sin \theta =\frac{1}{\sqrt{1+{(1+x)}^2}}$$$$\cos \left({\tan }^{-1}x\right)=\frac{1}{\sqrt{1+x^2}}$$$$1+{(1+x)}^2=1+x^2$$$$2x=-1\Rightarrow x=-\frac{1}{2}$$

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