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

Consider `f(x)=tan^(-1)(sqrt((1+sinx)/(1-sinx))), x in (0,pi/2)dot`A n

1.	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)`A. (0, 0)B. `(0, (2pi)/(3))`C. `((pi)/(6),0)`D. `((pi)/(4),0)`
Answer» Correct Answer - B 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)).`

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

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