The number of roots of (1−tanθ)(1+sin2θ)=1+tanθ for θ∈[0,2π] is
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a
3
b
4
c
5
d
none of these
answer is C.
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Detailed Solution
(1−tanθ)1+2tanθ/1+tan2θ=1+tanθ or (1−tanθ)(1+tanθ)2=(1+tanθ)1+tan2θ or (1+tanθ)1−tan2θ−1+tan2θ=0 or −2tan2θ=0,(1+tanθ)=0 or tanθ=0, or tanθ=−1⇒ θ=nπ or nπ−π/4,∀n∈Z, For θ∈[0,2π],θ=0,π,2π,3π/4,7π/4