यदि `(rcostheta-sqrt(3))^(2)+(rsintheta-1)^(2)=0` है तो `(rtantheta+sectheta)/(rsectheta+tantheta)` का मान क्या होगा?

Options

A. `(4)/(5)`
B. `(3)/(5)`
C. `(sqrt(3))/(4)`
D. `sqrt((5)/(3))`

Answer

a

Explanation

`(r cos theta-sqrt(3))^(2)+(r sin theta-1)^(2)=0`
`(r cos theta-sqrt(3))^(2)=0,(r sin theta-1)^(2)`
`=0r cos theta=sqrt(3)` ...........(i)
`r sin theta=1 ` ........(ii)
Squaring and adding equation
and ii
`r^(2)cos^(2)theta+r^(2)sin^(2)theta=3+1`
`r^(2)(cos^(2)theta+sin^(2)theta)=4`
`r^(2)=4`
r=2
`tantheta=(r sintheta)/(r cos theta)(1)/sqrt(3)` and
`r cos theta=sqrt(3)`
`cos theta=sqrt(3)/(r) sec theta=(r)/sqrt(3)`
`(r tan theta+sec theta)/(r sectheta+tan theta)=((r)/sqrt(3)+(r)/sqrt(3))/((r^(2))/sqrt(3)+(1)/sqrt(3))`
`=(r((2)/(sqrt(3))))/((r^(2)+1)/sqrt(3))=(2r)/(r^(2)+1)=(2xx2)/(2^(4)+1)=(4)/(6)`
Alternate
r=2
`tan theta=(r sin theta)/(r cos theta)=(1)/sqrt(3)`
`theta=30^(@)`
`=(2 tan 30^(@)+sec30^(@))/(2 sec30^(@)+tan30^(@))`
`=(2xx(1)/sqrt(3)+(2)/sqrt(3))/(2xx(2)/sqrt(3)+(1)/sqrt(3))=(4)/(5)`