SIN (x+y)\left\{ 1+\frac{dy}{dx} \right\}\] \[=-\frac{\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}\] \[\RIGHTARROW \sin (x+y)=1,\] so \[\cos (x+y)=0\] \[\therefore \] from \[(1)y=0\] and \[(x+y)=2n\pi +\frac{\pi }{2}\] TANGENT at \[\left( \frac{\pi }{2},0 \right)\] is \[x+2y=\frac{\pi }{2}\]

"> SIN (x+y)\left\{ 1+\frac{dy}{dx} \right\}\] \[=-\frac{\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}\] \[\RIGHTARROW \sin (x+y)=1,\] so \[\cos (x+y)=0\] \[\therefore \] from \[(1)y=0\] and \[(x+y)=2n\pi +\frac{\pi }{2}\] TANGENT at \[\left( \frac{\pi }{2},0 \right)\] is \[x+2y=\frac{\pi }{2}\]

">
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The equation of one of the tangents to the curve \[y=\cos (x+y),-2\pi \le x\le 2\pi \] that is parallel to the line \[x+2y=0,\] is

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[b] \[y=\cos (x+y)....(1)\] \[\therefore \frac{dy}{dx}=-\SIN (x+y)\left\{ 1+\frac{dy}{dx} \right\}\] \[=-\frac{\sin (x+y)}{1+\sin (x+y)}=-\frac{1}{2}\] \[\RIGHTARROW \sin (x+y)=1,\] so \[\cos (x+y)=0\] \[\therefore \] from \[(1)y=0\] and \[(x+y)=2n\pi +\frac{\pi }{2}\] TANGENT at \[\left( \frac{\pi }{2},0 \right)\] is \[x+2y=\frac{\pi }{2}\]

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Posted on 05 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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