If sec x = frac{25}{24} and x lies in the fir | vedclass

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

Question

(B) Given $\sec x = \frac{25}{24}$,we have $\cos x = \frac{24}{25}$.Since $x$ lies in the first quadrant,$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\f

Vedclass · Accepted Answer

$\frac{8}{5 \sqrt{2}}$

Vedclass · Answer

(B) Given $\sec x = \frac{25}{24}$,we have $\cos x = \frac{24}{25}$.Since $x$ lies in the first quadrant,$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{24}{25})^2} = \sqrt{\frac{625 - 576}{625}} = \sqrt{\frac{49}{625}} = \frac{7}{25}$.Now,consider $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 + \sin x$.Substituting the value of $\sin x$,we get $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = 1 + \frac{7}{25} = \frac{32}{25}$.Since $x$ is in the first quadrant,$0 Therefore,$\sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{\frac{32}{25}} = \frac{\sqrt{16 	imes 2}}{5} = \frac{4 \sqrt{2}}{5} = \frac{4 \sqrt{2} 	imes \sqrt{2}}{5 \sqrt{2}} = \frac{8}{5 \sqrt{2}}$.

If $\sec x = \frac{25}{24}$ and $x$ lies in the first quadrant,then $\sin \frac{x}{2} + \cos \frac{x}{2} =$

Similar Questions

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

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

$\operatorname{cosec} 2 \theta - \cot 2 \theta = ?$

$\sin 4\theta$ can be written as

If $k = \sin \frac{\pi}{18} \cdot \sin \frac{5\pi}{18} \cdot \sin \frac{7\pi}{18}$,then the numerical value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module