If y-log-1-sinx1-sinx- show that dydx-x

Given y=log√(1+sin x1 sin x)=12log(1+sin x1 sin x)=12log (1+sin x) 12log (1 sin x) Differentiating on both sides #160; d yd x=cos x2(1+sin x) cos x2( ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If y=log√1+...

Question

If $y = log \sqrt{\frac{1 + sin x}{1 - sin x}}$ , show that $\frac{d y}{d x} = sec x$

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Solution

Given $y = log \sqrt{\frac{1 + sin x}{1 - sin x}} = \frac{1}{2} log (\frac{1 + sin x}{1 - sin x}) = \frac{1}{2} log (1 + sin x) - \frac{1}{2} log (1 - sin x)$
Differentiating on both sides
$\frac{d y}{d x} = \frac{cos x}{2 (1 + sin x)} - \frac{cos x}{2 (1 - sin x)} = \frac{cos x}{2} \times \frac{1 - sin x + 1 + sin x}{1 - {sin}^{2} x} = \frac{cos x \times 2}{2 \times {cos}^{2} x} = sec x$

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If y=log√1+sinx1−sinx, show that dydx=secx

Given y=log√1+sinx1−sinx=12log(1+sinx1−sinx)=12log(1+sinx)−12log(1−sinx)Differentiating on both sides dydx=cosx2(1+sinx)−cosx2(1−sinx)=cosx2×1−sinx+1+sinx1−sin2x=cosx×22×cos2x=secx

If $y = log \sqrt{\frac{1 + sin x}{1 - sin x}}$ , show that $\frac{d y}{d x} = sec x$