If 0

2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2−1=32⇒ 2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2=12⇒4cos⁡x+y2cos⁡x−y2−4cos2⁡x+y2=1=cos2⁡x−y2+sin2⁡x−y2⇒4cos2⁡x+y2+cos2⁡x−y2−4cos⁡x+y2cos⁡x−y2+sin2⁡x−y2=0⇒cos⁡x−y2−2cos⁡x+y22+sin2⁡x−y2=0⇒ sin⁡x−y2=0⇒x=y⇒ cos⁡x−y2=2cos⁡x+y2  Now. put x=y ⇒ cos⁡x=12=cos⁡y∴ sin⁡x+cos⁡y=32+12=3+12