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Let $f (θ) = \frac{\cot θ}{1 - \tan θ} + \frac{\tan θ}{1 - \cot θ}, π < θ < \frac{3 π}{4}$ , then $f (\frac{9 π}{8})$ is equal to

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Correct option is B

f(θ)=1tan⁡θ1(1−tan⁡θ)+tan⁡θ⋅tan⁡θtan⁡θ−1=1−tan3⁡θtan⁡θ(1−tan⁡θ)=1+tan⁡θ+tan2⁡θtan⁡θ=1+2sin⁡2θ∴f9π8=1+2sin⁡9π4=1+22

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Let f(θ)=cot⁡θ1−tan⁡θ+tan⁡θ1−cot⁡θ,π<θ<3π4 , then f9π8 is equal to

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Let $f (θ) = \frac{\cot θ}{1 - \tan θ} + \frac{\tan θ}{1 - \cot θ}, π < θ < \frac{3 π}{4}$ , then $f (\frac{9 π}{8})$ is equal to