Consider f(x) = tan-1 (√(1+ sin x/1- sin x)) , x ϵ (0, (π/2)) A normal to y = f(x) at x = (π/2) also passes through the point :

Question

Consider f(x) = tan-1 (√(1+ sin x/1- sin x)) , x ϵ (0, (π/2)) A normal to y = f(x) at x = (π/2) also passes through the point : (A) (0, 0) (B) (0, (2 π/3)) (C) ( (π/6) , 0 ) (D) ( (π/4) , 0 )

Tardigrade Admin · Accepted Answer

f(x) = tan-1 tan ((π/4) + (x/2)) = (π/4) + (x/2) If x = (π/6) , f (π/6) = (π/4) + (π/12) = (4π/12) = (π/3) f'(x) = (1/2) Normal y - (π/3) =- 2 ( x - (π/6)) So, (0, (2π/3)) Satisfy

Q. Consider $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x ϵ (0, \frac{π}{2})$ A normal to $y = f (x)$ at $x = \frac{π}{2}$ also passes through the point :

(0, 0)

$(0, \frac{2 π}{3})$

$(\frac{π}{6}, 0)$

$(\frac{π}{4}, 0)$

$f (x) = tan^{- 1} tan (\frac{π}{4} + \frac{x}{2}) = \frac{π}{4} + \frac{x}{2}$
If $x = \frac{π}{6}, f \frac{π}{6} = \frac{π}{4} + \frac{π}{12} = \frac{4 π}{12} = \frac{π}{3}$
$f^{'} (x) = \frac{1}{2}$
Normal $y - \frac{π}{3} = - 2 (x - \frac{π}{6})$
So, $(0, \frac{2 π}{3})$ Satisfy

Q. Consider f(x)=tan−1(1−sinx1+sinx​​),xϵ(0,2π​) A normal to y=f(x) at x=2π​ also passes through the point :

(0, 0)

(0,32π​)

(6π​,0)

(4π​,0)

f(x)=tan−1tan(4π​+2x​)=4π​+2x​ If x=6π​,f6π​=4π​+12π​=124π​=3π​ f′(x)=21​ Normal y−3π​=−2(x−6π​) So, (0,32π​) Satisfy

Q. Consider $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x ϵ (0, \frac{π}{2})$ A normal to $y = f (x)$ at $x = \frac{π}{2}$ also passes through the point :

$(0, \frac{2 π}{3})$

$(\frac{π}{6}, 0)$

$(\frac{π}{4}, 0)$

$f (x) = tan^{- 1} tan (\frac{π}{4} + \frac{x}{2}) = \frac{π}{4} + \frac{x}{2}$
If $x = \frac{π}{6}, f \frac{π}{6} = \frac{π}{4} + \frac{π}{12} = \frac{4 π}{12} = \frac{π}{3}$
$f^{'} (x) = \frac{1}{2}$
Normal $y - \frac{π}{3} = - 2 (x - \frac{π}{6})$
So, $(0, \frac{2 π}{3})$ Satisfy