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Question Number 201427 by mathlove last updated on 06/Dec/23

$$\{\begin{array}{l}sin(x+y)=cos(x-y)\\ tanx-tany=1\end{array}$$$$(x,y)=(?,?)$$

Answered by Rasheed.Sindhi last updated on 06/Dec/23

$$\{\begin{array}{l}sin(x+y)=cos(x-y)...\left(i\right)\\ tanx-tany=1.........\left(ii\right)\end{array}$$$$(x,y)=(?,?)$$$$\left(i\right)\Rightarrow x+y=\left(\pi /2\right)-(x-y)$$$$2x=\pi /2$$$$x=\pi /4$$$$\left(ii\right)\Rightarrow \tan \left(\pi /4\right)-\tan y=1$$$$\Rightarrow 1-\tan y=1$$$$\Rightarrow \tan y=0$$$$\Rightarrow y=0$$$$\mathcal{G}enerally(x,y)=(\frac{\pi }{4}+2n\pi ,2n\pi )$$

Commented by mathlove last updated on 06/Dec/23

$$thanks$$

Answered by mr W last updated on 06/Dec/23

$$\sin (x+y)=\cos (x-y)=\sin (\frac{\pi }{2}-x+y)$$$$k\pi +{(-1)}^k(x+y)=\frac{\pi }{2}-x+y$$$$\Rightarrow \frac{(2k-1)\pi }{2}+[1+{(-1)}^k]x=[1-{(-1)}^k]y$$$$case1:k=2n$$$$\frac{(4n-1)\pi }{2}+2x=0$$$$\Rightarrow x=-n\pi +\frac{\pi }{4}$$$$\tan y=\tan x-1=1-1=0$$$$\Rightarrow y=m\pi$$$$case2:k=2n+1$$$$\frac{(4n+1)\pi }{2}=2y$$$$\Rightarrow y=n\pi +\frac{\pi }{4}$$$$\tan x=1+\tan y=1+1=2$$$$\Rightarrow x=m\pi +{\tan }^{-1}2$$$$$$$$summary(withn,m\in Z):$$$$\{\begin{array}{l}x=n\pi +\frac{\pi }{4}\\ y=m\pi \end{array}or\{\begin{array}{l}x=n\pi +{\tan }^{-1}2\\ y=m\pi +\frac{\pi }{4}\end{array}$$

Commented by mathlove last updated on 06/Dec/23

$$thanks$$

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