Q.
15[tan2θ+sin2θ]+8=0 if
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a
tanθ=1/2
b
sinθ=1/4
c
tanθ=2
d
cosθ=1/5
answer is C.
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Detailed Solution
We have tan2θ+sin2θ=−8/15⇒ 2tanθ1−tan2θ+2tanθ1+tan2θ=−815⇒ 15×4tanθ+81−tan4θ=0⇒ 8tan4θ+60tanθ−8=0which is satisfied if tanθ=2
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