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

a
125
b
– 3328
c
33100
d
1213

detailed solution

Correct option is C

tan⁡θ(3tan⁡θ+3)=(2+2tan⁡θ)2⇒(1+tan⁡θ)(4+tan⁡θ)=0⇒tan⁡θ=−4 as tan⁡θ≠−1 So 7−5cot⁡θ9−4sec2⁡θ−1=7+5/49+16=33100

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