If tan theta = frac{1}{sqrt{5}} and theta lie | vedclass

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(A) Given that $	an 	heta = \frac{1}{\sqrt{5}}$.We know that $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}}$.Let the opposite side be $1$ an

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$\sqrt{\frac{5}{6}}$

Vedclass · Answer

(A) Given that $	an 	heta = \frac{1}{\sqrt{5}}$.We know that $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}}$.Let the opposite side be $1$ and the adjacent side be $\sqrt{5}$.Using the Pythagorean theorem,the hypotenuse is $\sqrt{1^2 + (\sqrt{5})^2} = \sqrt{1 + 5} = \sqrt{6}$.Since $	heta$ lies in the first quadrant,all trigonometric ratios are positive.Therefore,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{\sqrt{5}}{\sqrt{6}} = \sqrt{\frac{5}{6}}$.

If $\tan \theta = \frac{1}{\sqrt{5}}$ and $\theta$ lies in the first quadrant,the value of $\cos \theta$ is

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