Period tanθ(1+sec2θ) (1+sec4θ)(1+sec8θ) is

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π2

2π

π8

answer is D.

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Detailed Solution

tanθ1+sec2θ(1+sec4θ)(1+sec8θ) =tanθ1+cos2θcos2θ1+cos4θcos4θ1+cos8θcos8θ =sinθcosθ2cos2θcos2θ2cos22θcos4θ2cos24θcos8θ =sin8θcos8θ=tan8θ period is π8

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