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Question Number 63643 by vajpaithegrate@gmail.com last updated on 06/Jul/19

$$\mathrm{P}(\alpha ,\beta )\mathrm{Q}(\gamma ,\delta )\mathrm{are}\mathrm{two}\mathrm{points}\mathrm{lie}\mathrm{on}\mathrm{curve}$$$${\tan }^2(\mathrm{x}+\mathrm{y})+{\cos }^2(\mathrm{x}+\mathrm{y})+{\mathrm{y}}^2+2\mathrm{y}=0\mathrm{on}$$$$\mathrm{XY}\mathrm{plane}.\mathrm{If}\mathrm{d}=\mathrm{PQ}\mathrm{then}\cos \mathrm{d}=$$$$\mathrm{ans}:\pm 2\mathrm{n}\pi ,\mathrm{n}\in \mathrm{N}$$

Answered by MJS last updated on 06/Jul/19

$$\frac{d}{dy}[y^2+2y]=2y+2=0\Rightarrow y=-1\Rightarrow$$$$\Rightarrow y^2+2y\mathrm{has}\mathrm{its}\mathrm{minimum}\mathrm{at}\left(\begin{array}{c}-1\\ -1\end{array}\right)$$$$\frac{d}{dt}[{\tan }^{2}t+{\cos }^{2}t]=2\frac{\tan t}{{\cos }^{2}t}-2\sin t\cos t=0\Rightarrow$$$$\Rightarrow t=n\pi$$$$\Rightarrow {\tan }^{2}t+{\cos }^{2}t\mathrm{has}\mathrm{its}\mathrm{minima}\mathrm{at}\left(\begin{array}{c}n\pi \\ 1\end{array}\right)$$$$\Rightarrow \mathrm{the}\mathrm{only}\mathrm{possibilities}\mathrm{for}\mathrm{pairs}\left(\begin{array}{c}x\\ y\end{array}\right)\mathrm{are}\left(\begin{array}{c}n\pi +1\\ -1\end{array}\right)$$$$\mathrm{two}\mathrm{points}\left(\begin{array}{c}m\pi +1\\ -1\end{array}\right)\mathrm{and}\left(\begin{array}{c}n\pi +1\\ -1\end{array}\right)\mathrm{have}\mathrm{the}$$$$\mathrm{distance}d=\mid m-n\mid \pi =k\pi ;m,n\in \mathbb{Z},k\in \mathbb{N}$$$$\Rightarrow \cos d=\pm 1$$

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