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Q.

The general solution of the equation 2cotθ2=1+cotθ2 is

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a

nπ+−1nπ4,n∈Z

b

nπ+−1nπ3,n∈Z

c

nπ+−1nπ6,n∈Z

d

nπ2

answer is C.

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Detailed Solution

We have 2cotθ2=1+cotθ2⇒2×2cosθ2×cosθ22sinθ2×cosθ2=1+cotθ2⇒21+cosθsinθ=cosec2θ+2cotθ ⇒2cosecθ+2cotθ=cosec2θ+2cotθ⇒cosecθ=2⇒sinθ=12⇒θ=nπ+−1nπ6,n∈Z
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