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If $\tan θ, 2 \tan θ + 2 and 3 \tan θ + 3$ are in GP, then the value of $\frac{7 - 5 \cot θ}{9 - 4 \sqrt{\sec^{2} θ - 1}}$ is

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a

\frac{12}{5}

b

- \frac{33}{28}

c

\frac{33}{100}

d

\frac{12}{13}

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detailed solution

Correct option is C

$\begin{array}{l} Since, (2 \tan θ + 2)^{2} = \tan θ (3 \tan θ + 3) \\ \Rightarrow 4 \tan^{2} θ + 8 \tan θ + 4 = 3 \tan^{2} θ + 3 \tan θ \\ \Rightarrow \tan^{2} θ + 5 \tan θ + 4 = 0 \\ \Rightarrow \tan θ + 4) (\tan θ + 1) = 0 \\ \Rightarrow 7 - 5 \cot θ \\ \Rightarrow 7 + \frac{5}{4} = \frac{3}{9 - 4 (- 4)} = \frac{33}{100} \end{array}$

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