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If $\sec x + \sec^{2} x = 1$ then the value of $\tan^{8} x - \tan^{4} x - 2 \tan^{2} x + 1$ is equal to

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Correct option is C

We have, sec⁡x+sec2⁡x=1 sec⁡x=−sec2⁡x−1=−tan2⁡x⇒sec2⁡x=tan4⁡x⇒1+tan2⁡x=tan4⁡x⇒1+tan2⁡x2=tan8⁡x⇒1+2tan2⁡x+tan4⁡x=tan8⁡x⇒tan8⁡x−tan4⁡x−2tan2⁡x=1⇒tan8⁡x−tan4⁡x−2tan2⁡x+1=2

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

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If $\sec x + \sec^{2} x = 1$ then the value of $\tan^{8} x - \tan^{4} x - 2 \tan^{2} x + 1$ is equal to