Question

under root 1+sinx / 1-sinx. differentiate

Answer 1

$\sqrt{ \frac{1 + \sin(x) }{1 - \sin(x) } } \\ = \sqrt{ \frac{1 + \sin(x) }{1 - \sin(x) } \times \frac{1 + \sin(x) }{1 + \sin(x) } } \\ = \sqrt{ \frac{(1 + \sin(x)) ^{2} }{1 - \sin ^{2} (x) } } = \frac{1 + \sin(x) }{ \cos(x) } \\ = \sec(x) + \tan(x) \\ now \: \: \: differentiating.... \\ \frac{d}{dx} ( \sec(x) + \tan(x) ) \\ = \sec(x) \tan(x) + \sec ^{2} (x) ...ans$
Hope this is ur required answer.

Answer 2

Answer:

Step-by-step explanation :

y = √(1+sin x)/(1-sin x)

rationalising numerator and denominator

y = √(1+sin x)/(1-sin x)*(1+sin x)/(1-sin x)

y = √(1+sin x)^2 / (1-sin^2x)

y = √(1+sin x)^2 / cos^2x

y = (1+sin x) / cos x

differentiating with respect to x

d/dx y = d/dx (1+sin x)/cos x

dy/dx = [cos x d/dx (1+sin x) - (1+sin x) d/dx cos x] / (cos x)^2

d/dx = [cos x . cos x - (1+sin x)(-sin x)] / cos^2x

d/dx = [cos^2x + sinx + sin^2x] / cos^2x

d/dx = (1+sin x)/cos^2x

d/dx = sec^2x + sec x . tan x .........(ans)