If sin−1⁡x2+2x+1+sec−1⁡x2+2x+1=π2,x≠0 then the value of 2sec−1⁡x2+sin−1⁡x2 is equal to

We have,sin−1⁡x2+2x+1+sec−1⁡x2+2x+1=π2sin−1⁡x2+2x+1+cos−1⁡1x2+2x+1=π2⇒x2+2x+1=1x2+2x+1⇒x2+2x+1=1⇒x=0,−2.For x=0, we find that sec−1⁡x2 is not defined.For x=-2, we have2sec−1⁡x2+sin−1⁡x2=2sec−1⁡(−1)+sin−1⁡(−1)=2×π−π2=3π2