Q.

15[tan⁡2θ+sin⁡2θ]+8=0 if

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a

tan⁡θ=1/2

b

sin⁡θ=1/4

c

tan⁡θ=2

d

cos⁡θ=1/5

answer is C.

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Detailed Solution

We have tan⁡2θ+sin⁡2θ=−8/15⇒ 2tan⁡θ1−tan2⁡θ+2tan⁡θ1+tan2⁡θ=−815⇒ 15×4tan⁡θ+81−tan4⁡θ=0⇒ 8tan4⁡θ+60tan⁡θ−8=0which is satisfied if tan⁡θ=2
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