Consider `f(x) =tan^(-1)(sqrt((1+sinx)/(1-sinx))),c in(0,(pi)/(2)).` A normal to `y=(x) at x=(pi)/(6)` also pasess through the point

Options

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

Answer

B

Explanation

We have,
`y ="tan" ^(-1)sqrt((1+sinx)/(1-sinx))= tan^(-1)(("cos"(x)/(2) + " sin" (x)/(2))/("cos"(x)/(2) - " sin" (x)/(2)))`
` rArr y=tan^(-1) (tan((x)/(4)+(x)/(2)))`
` rArr y= (pi)/(4)+ (x)/(2)`
` rArr (dy)/(dx) = (1)/(2) rArr (1)/(dy//dx)= -2.`
When `x=(pi)/(6),y=(pi)/(4)+(pi)/(12)=(pi)/(3).`
The equation of the normal at `(pi//6, pi//3)` is
` y-(pi)/(3)= -2(x-(pi)/(6)) or , 2x+y-(2pi)/(3) =0`
Clearly, it passes through `(0, (2pi)/(3)).`