If (1+tan ^{2} theta)=frac{625}{49} and theta | vedclass

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Question

(C) Given $(1+	an ^{2} 	heta)=\frac{625}{49}$.Using the trigonometric identity $1+	an ^{2} 	heta=\sec ^{2} 	heta$,we have $\sec ^{2} 	heta=\frac

Vedclass · Accepted Answer

$\frac{\sqrt{31}}{5}$

Vedclass · Answer

(C) Given $(1+	an ^{2} 	heta)=\frac{625}{49}$.Using the trigonometric identity $1+	an ^{2} 	heta=\sec ^{2} 	heta$,we have $\sec ^{2} 	heta=\frac{625}{49}$.Taking the square root,$\sec 	heta=\frac{25}{7}$ (since $	heta$ is acute,$\sec 	heta$ is positive).Therefore,$\cos 	heta=\frac{1}{\sec 	heta}=\frac{7}{25}$.Using $\sin ^{2} 	heta + \cos ^{2} 	heta = 1$,we get $\sin 	heta=\sqrt{1-\cos ^{2} 	heta} = \sqrt{1-\left(\frac{7}{25}ight)^{2}}$.$\sin 	heta = \sqrt{1-\frac{49}{625}} = \sqrt{\frac{625-49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}$.Now,calculate $\sqrt{\sin 	heta+\cos 	heta} = \sqrt{\frac{24}{25}+\frac{7}{25}}$.$= \sqrt{\frac{31}{25}} = \frac{\sqrt{31}}{5}$.

If $(1+\tan ^{2} \theta)=\frac{625}{49}$ and $\theta$ is acute,then what is the value of $\sqrt{\sin \theta+\cos \theta}$?

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