Find the values of trignometric functions in Questions 6 to 10.
`sin765^(@)`

Answer

b

Explanation

Given, then `cosx=-1/2`, and x lies in third quadrant.
`therefore sin^(2)+cos^(2)x=1`
`therefore sin^(2)x+(-1/2)^(2)=1 rArr sin^(2)x+1/4=1`
`rArr sin^(2)x=1-1/4=3/4`
`rArr sinx=+-sqrt(3)/2`
but x lins in third quadrant.
Therefore, `sinx=-sqrt(3)/2`
then `sinx=-sqrt(3)/2`
then `tanx=(sinx)/(cosx) = (-sqrt(3)/2)/(-1/2)`
`=sqrt(-3)/2 xx -2/1 = sqrt(3)`
`therefore tanx=sqrt(3)`
`therefore cotx=1/(tanx) = sqrt(3)`
`therefore secx=1/(cosx)`
`therefore secx=1/(-1/2) = 1 xx -2/1=-2`
`therefore secx=-2`
`therefore "cosec "x=1/(sinx)`
`therefore "cosec"x=1/(sinx)`
`therefore "cosec" x =1/(-sqrt(3)/2) = 1 xx -2/sqrt(3) = -2/sqrt(3)`
`sinx=-sqrt(3)/2`,
then `tanx=(sinx)/(cosx) = (-sqrt(3)/2)/(-1/2)`
`=-sqrt(3)/2 xx -2/1= sqrt(3)`
`therefore tanx=sqrt(3)`
`cotx=1/(tanx) rArr cotx = 1/sqrt(3)`
`therefore secx=1/(cosx)`
`therefore secx=1/(-1/2) = 1 xx -2/1=-2`
`secx=-2`
`therefore "cosec "x=1/(sinx)`
`therefore "cosec"x=1/(-sqrt(3)/2) =1 xx -2/sqrt(3) = -2/sqrt(3)`
`sinx=-sqrt(3)/(2), tanx=sqrt(3)`,
`cotx=1/sqrt(3), secx=-2`
and `"cosec "x=-2sqrt(2)`. Ans.