If f(x) = (1-cos x/1-sin x), then f' ((π/2)) is equal to

Question

If f(x) = (1-cos x/1-sin x), then f' ((π/2)) is equal to (A) 1 (B) 0 (C) ∞ (D) does not exist

Tardigrade Admin · Accepted Answer

f(x) = ((1-sin x)(sin x)-(1-cos x)(-cos x)/(1-sin x)2) = (sin x - sin2 x+cos x-cos2 x/(1-sin x)2) = (sin x+cos x - 1/(1-sin x)2) ∴ f'((π/2)) = (1+0-1/(1-1)2) = (0/0) Therefore, f'((π /2)) does not exist.

Q. If $f (x) = \frac{1 - cos x}{1 - s in x}$ , then $f^{'} (\frac{π}{2})$ is equal to

$1$

$0$

$\infty$

does not exist

Q. If f(x)=1−sinx1−cosx​, then f′(2π​) is equal to

1

0

∞

does not exist

f(x)=(1−sinx)2(1−sinx)(sinx)−(1−cosx)(−cosx)​ =(1−sinx)2sinx−sin2x+cosx−cos2x​=(1−sinx)2sinx+cosx−1​ ∴f′(2π​)=(1−1)21+0−1​=00​ Therefore, f′(2π​) does not exist.

Q. If $f (x) = \frac{1 - cos x}{1 - s in x}$ , then $f^{'} (\frac{π}{2})$ is equal to

$1$

$0$

$\infty$