If tan ⁡θ,2tan ⁡θ+2,3tan ⁡θ+3 are in G.P then the value of 7−5cot⁡θ9−4sec2⁡θ−1is

(2tan⁡θ+2)2=tan⁡θ(3tan⁡θ+3)⇒tan2⁡θ+5tan⁡θ+4=0⇒tan⁡θ=−1,−4 but tan⁡θ=−4; cotθ=-14; 7−5cot⁡θ9−4sec2⁡θ−1=7+549-4(4)=-3328

If tan ⁡θ,2tan ⁡θ+2,3tan ⁡θ+3 are in G.P then the value of 7−5cot⁡θ9−4sec2⁡θ−1is

(2tan⁡θ+2)2=tan⁡θ(3tan⁡θ+3)⇒tan2⁡θ+5tan⁡θ+4=0⇒tan⁡θ=−1,−4 but tan⁡θ=−4; cotθ=-14; 7−5cot⁡θ9−4sec2⁡θ−1=7+549-4(4)=-3328

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

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-33/28

12/5

33/100

12/13

answer is B.

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Detailed Solution

$\begin{array}{l} (2 \tan θ + 2)^{2} = \tan θ (3 \tan θ + 3) \\ \Rightarrow \tan^{2} θ + 5 \tan θ + 4 = 0 \Rightarrow \tan θ = - 1, - 4 \\ but \tan θ = - 4; c o t θ = - \frac{1}{4}; \frac{7 - 5 \cot θ}{9 - 4 \sqrt{\sec^{2} θ - 1}} = \frac{7 + \frac{5}{4}}{9 - 4 (4)} = - \frac{33}{28} \end{array}$