If y(x)= cot -1((√1+ sin x+√1- sin x/√1+ sin x-√1- sin x)), x ∈((π/2), π) then ( dy / dx ) at x =(5 π/6) is:

Question

If y(x)= cot -1((√1+ sin x+√1- sin x/√1+ sin x-√1- sin x)), x ∈((π/2), π) then ( dy / dx ) at x =(5 π/6) is: (A) -(1/2) (B) -1 (C) (1/2) (D) 0

Tardigrade Admin · Accepted Answer

y(x)= cot -1[( cos (x/2)+ sin (x/2)+ sin (x/2)- cos (x/2)/ cos (x)2+ sin (x)2- sin (x/2)+ cos (x/2)] y(x)= cot /-1)( tan (x/2))=(π/2)-(x/2) y prime(x)=(-1/2)

Q. If $y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x}), x \in (\frac{π}{2}, π)$ then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:

$- \frac{1}{2}$

$- 1$

$\frac{1}{2}$

$0$

$y (x) = cot^{- 1} [\frac{c o s \frac{x}{2} + s i n \frac{x}{2} + s i n \frac{x}{2} - c o s \frac{x}{2}}{c o s \frac{x}{2} + s i n \frac{x}{2} - s i n \frac{x}{2} + c o s \frac{x}{2}}]$
$y (x) = cot^{- 1} (tan \frac{x}{2}) = \frac{π}{2} - \frac{x}{2}$
$y^{'} (x) = \frac{- 1}{2}$

Q. If y(x)=cot−1(1+sinx​−1−sinx​1+sinx​+1−sinx​​),x∈(2π​,π) then dxdy​ at x=65π​ is:

−21​

−1

21​

0

y(x)=cot−1[cos2x​+sin2x​−sin2x​+cos2x​cos2x​+sin2x​+sin2x​−cos2x​​] y(x)=cot−1(tan2x​)=2π​−2x​ y′(x)=2−1​

Q. If $y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x}), x \in (\frac{π}{2}, π)$ then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:

$- \frac{1}{2}$

$- 1$

$\frac{1}{2}$

$0$

$y (x) = cot^{- 1} [\frac{c o s \frac{x}{2} + s i n \frac{x}{2} + s i n \frac{x}{2} - c o s \frac{x}{2}}{c o s \frac{x}{2} + s i n \frac{x}{2} - s i n \frac{x}{2} + c o s \frac{x}{2}}]$
$y (x) = cot^{- 1} (tan \frac{x}{2}) = \frac{π}{2} - \frac{x}{2}$
$y^{'} (x) = \frac{- 1}{2}$