Consider `f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot` A normal to `y=f(x)` at `x=pi/6` also passes through the point: (1) (0, 0) (2) `(0,(2pi)/3)` (3) `(pi/6,0)` (4) `(pi/4,0)`

Options

A. `(0, 0)`
B. `(0, (2pi)/(3))`
C. `((pi)/(6), 0)`
D. `((pi)/(4), 0)`

Answer

B

Explanation

We have, `f(x) = tan ^(-1) sqrt((1 + sin x)/( 1-sin x)),x in (0, (pi)/(2))`
`rArr " " f(x) = tan ^(-1) sqrt(((cos""(x)/(2) + sin ""(x)/(2))^(2)) /(( cos ""(x)/(2) - sin ""(x)/(2))^(2)))`
`" " = tan ^(-1) ((cos""(x)/(2) + sin ""(x)/(2))/(cos ""(x)/(2) - sin ""(x)/(2)))`
`" " ( because cos""(x)/(2) gt sin ""(x)/(2)" for " 0 lt (x)/(2) lt (pi)/(4))`
`" " = tan^(-1) ((1+ tan""(x)/(2))/( 1- tan ""(x)/(2)))`
`= tan ^(-1) [ tan ((pi)/(4) + (x)/(2))] = (pi)/(4) + (x)/(2)`
`rArr f'(x) = (1)/(2) rArr f' ((pi)/(6)) = (1)/(2)`
Now, equation of normal at `x = (pi)/(6)` is given by
`(y - f((pi)/(6)))=-2 (x - (pi)/(6))`
`rArr (y - (pi)/(3)) =- 2 (x - (pi)/(6)) [ because f((pi)/(6)) = (pi)/(4) + (pi)/( 12) = ( 4pi)/(12) = (pi)/(3)] `
which passes through `(0, (2pi)/(3))`.