If `sintheta_1sintheta_2-costheta_1costheta_2+1=0,` then the value of `tan((theta_1)/2)cot((theta_2)/2)` is equal to `-1` (b) 1 (c) 2 (d) `-2`

Options

A. `-1`
B. 1
C. 2
D. `-2`

Answer

A

Explanation

`sin theta_(1) sintheta_(2)=-1`
or `cos (theta_(1)+theta_(2))=1`
`rArr theta_(1)+theta_(2)=2npi, n in I`
or `(theta_(1))/(2)+(theta_(2))/(2)=npi`
Thus, `tan""(theta_(1))/(2)cot""(theta_(2))/(2)=tan""(theta_(1))/(2)cot(npi-(theta_(1))/(2))`
`=-tan""(theta_(1))/(2)cot""(theta_(1))/(2)=-1`