If sec⁡x+sec2⁡x=1 then the value of tan8⁡x−tan4⁡x−2tan2⁡x+1 will be equal to

sec⁡x=1−sec2⁡x⇒ sec⁡x=−tan2⁡xSquaring both side, we get sec2⁡x=tan4⁡x⇒ 1+tan2⁡x=tan4⁡xSquaring both side, we get tan8⁡x=tan4⁡x+2tan2⁡x+1⇒ tan8⁡x−tan4⁡x−2tan2⁡x=1⇒ tan8⁡x−tan4⁡x−2tan2⁡x+1=2