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Q.

If tanθ+3cotθ=5secθ , then θ=(nz)  

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a

nπ+(1)nπ4,nz

b

nπ+(1)nπ2ornπ+(1)nπ6,nz

c

nπ+(1)nπ6ornπ+(1)nπ3,nz

d

nπ+(1)nπ6,nz

answer is D.

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Detailed Solution

tanθ+3cotθ=5secθ

sinθcosθ+3.cosθsinθ=5cosθ

sin2θ+3cos2θ=5sinθ

2sin2+5sinθ3=0

(2sinθ1)(sinθ+3)=0

sinθ=3  is not possible

sinθ=12  

θ=π6

 G.S is θ=nπ+(1)nπ6

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