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Q.

If sin⁡θ1sin⁡θ2−cos⁡θ1cos⁡θ2+1=0then the value of tan⁡θ1/2cot⁡θ2/2 is equal to

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a

-1

b

1

c

2

d

-2

answer is A.

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Detailed Solution

sin⁡θ1sin⁡θ2−cos⁡θ1cos⁡θ2=−1 or cos⁡θ1+θ2=1⇒ θ1+θ2=2nπ,n∈1 or  θ12+θ22=nπ Thus, tan⁡θ12cot⁡θ22=tan⁡θ12cot⁡nπ−θ12           =−tan⁡θ12cot⁡θ12=−1
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