Question

If 2 `sin(theta+(pi)/(3))=cos(theta-(pi)/(6))`, then `tantheta`=
A. `sqrt(3)`
B. `-(1)/(sqrt(3))`
C. `(1)/(sqrt(3))`
D. `-sqrt(3)`

Answer 1

Correct Answer - D
We have,
`2sin(theta+(pi)/(3))=cos(theta=(pi)/(6))`
`rArr2[sintheta"cos"(pi)/(3)+costheta"sin"(pi)/(3)]`
`=costheta"cos"(pi)/(6)+sintheta"sin"(pi)/(6)`
`rArr2[(sintheta)/(2)+(sqrt(3)costheta)/(2)]=(sqrt(3))/(2)costheta+(1)/(2)sintheta`
`rArr2sintheta+2sqrt(3)costheta=sqrt(3)costheta+sintheta`
`rArrsintheta=-sqrt(3)costheta`
`rArrtantheta=-sqrt(3)`