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

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(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

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

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(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 $\sin \frac{x}{2} + \cos \frac{x}{2} =$

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