frac{sqrt{1 + sin x} + sqrt{1 - sin x}}{sqrt{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression: $E = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}$
Since $1 \pm \sin x = \cos^2 \frac{x}

Vedclass · Accepted Answer

$\cot \frac{x}{2}$

Vedclass · Answer

(B) Given expression: $E = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}$ Since $1 \pm \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \pm 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} \pm \sin \frac{x}{2})^2$,we have $\sqrt{1 \pm \sin x} = |\cos \frac{x}{2} \pm \sin \frac{x}{2}|$. For $x$ in the $II^{nd}$ quadrant,$\frac{\pi}{2} < x < \pi$,so $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$. In this interval,$\cos \frac{x}{2} > 0$ and $\sin \frac{x}{2} > 0$,with $\cos \frac{x}{2} > \sin \frac{x}{2}$. Thus,$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$. Substituting these into the expression: $E = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}$ $E = \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}$. Note: The provided option $(b)$ $ an \frac{x}{2}$ is incorrect based on the quadrant analysis. The correct result is $\cot \frac{x}{2}$.

$\frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = $ (when $x$ lies in $II^{nd}$ quadrant)

Similar Questions

If $\sin x - \sin y = \frac{1}{2}$ and $\cos x - \cos y = 1$,then $\tan(x + y)$ is equal to

If $\tan \frac{\theta}{2} = t$,then $\frac{1 - t^2}{1 + t^2}$ is equal to

If $x + \frac{1}{x} = 2 \cos \theta,$ then $x^3 + \frac{1}{x^3} = $

If $2\cos^2 \theta - 2\sin^2 \theta = 1$,then $\theta = \dots ^\circ$.

$\frac{1-\tan^2 15^{\circ}}{1+\tan^2 15^{\circ}} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module