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Question Number 170968 by 281981 last updated on 05/Jun/22

Commented by mr W last updated on 05/Jun/22

$${\tan }^2(x+y)+{\cos }^2(x+y)-1+{(y+1)}^2=0$$$${\tan }^2(x+y)-{\sin }^2(x+y)+{(y+1)}^2=0$$$$\frac{{\sin }^4(x+y)}{{\cos }^2(x+y)}+{(y+1)}^2=0$$$$\Rightarrow \sin (x+y)=0\cap y+1=0$$$$\Rightarrow x+y=k\pi \cap y+1=0$$$$\Rightarrow x=k\pi +1\cap y=-1$$$$d=PQ=(k_1\pi +1)-(k_2\pi +1)=n\pi$$$$\cos d=\cos \left(n\pi \right)={(-1)}^n$$

Commented by 281981 last updated on 05/Jun/22

$$\mathbf{tnq}\mathbf{sir}$$

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