tan(π4+θ)⋅tan(3π4+θ)=

tan(π4+θ)⋅tan(3π4+θ) [tanπ4+tanθ1−tanπ4⋅tanθ][tan3π4+tanθ1−tan3π4⋅tanθ] [∵tan(A+B)=tanA+tanB1−tanA⋅tanB] =1+tanθ1−tanθ×−1+tanθ1+tanθ =(1+tanθ)(1−tanθ)×−(1−tanθ)(1+tanθ) =−1

tan(π4+θ)⋅tan(3π4+θ)=

tan(π4+θ)⋅tan(3π4+θ) [tanπ4+tanθ1−tanπ4⋅tanθ][tan3π4+tanθ1−tan3π4⋅tanθ] [∵tan(A+B)=tanA+tanB1−tanA⋅tanB] =1+tanθ1−tanθ×−1+tanθ1+tanθ =(1+tanθ)(1−tanθ)×−(1−tanθ)(1+tanθ) =−1

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$\tan (\frac{π}{4} + θ) \cdot \tan (\frac{3 π}{4} + θ) =$

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\frac{1}{\sqrt{3}}

answer is C.

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Detailed Solution

\tan (\frac{π}{4} + θ) \cdot \tan (\frac{3 π}{4} + θ)

$[\frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} \cdot \tan θ}] [\frac{\tan \frac{3 π}{4} + \tan θ}{1 - \tan \frac{3 π}{4} \cdot \tan θ}]$ $[∵ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}]$