Question

If $y=\frac{\sqrt{1+\mathrm{sinx}}+\sqrt{1-\mathrm{sinx}}}{\sqrt{1+\mathrm{sinx}}-\sqrt{1-\mathrm{sinx}}}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{2}\cos e{c}^2\frac{x}{2}$
• B. $\frac{1}{2}\cos ec\frac{x}{2}$
• C. $\frac{1}{2}\cos e{c}^{2}x$
• D. $\cos e{c}^2\frac{x}{2}$

Answer 1

The correct option is A

$\frac{1}{2}\cos e{c}^2\frac{x}{2}$

Explanation for the correct option

Step 1: Simplify

Given, $\mathrm{y}=\frac{\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}}}{\sqrt{1-\mathrm{sinx}}-\sqrt{1+\mathrm{sinx}}}$

Multiply numerator and denominator with $\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}}$.

$\begin{array}{rcl}y& =& \frac{(\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}})\left(\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}}\right)}{(\sqrt{1-\mathrm{sinx}}-\sqrt{1+\mathrm{sinx}})\left(\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}}\right)}\\ & =& \frac{{(\sqrt{1-\mathrm{sinx}}+\sqrt{1+\mathrm{sinx}})}^2}{-2\sin x}\\ & =& \frac{2(1+\cos x)}{-2\sin x}\\ & =& \frac{-2{\cos }^2{\displaystyle \frac{x}{2}}}{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}\\ & =& -cot\frac{x}{2}\end{array}$

Step2: Find the $\frac{dy}{dx}$

Since, $y=-cot\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Differentiate with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=-\left(\cos e{c}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right) \\ =\frac{1}{2}\cos e{c}^2\frac{x}{2} \end{gathered}$$

Hence, A is the correct option.