If tan $x = - \frac{1}{\sqrt{5}}$ and θ lies in the IV quadrant, then the value of cos x is
(a) $\frac{\sqrt{5}}{\sqrt{6}}$

(b) $\frac{2}{\sqrt{6}}$

(c) $\frac{1}{2}$

(d) $\frac{1}{\sqrt{6}}$

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Solution

(a) $\frac{\sqrt{5}}{\sqrt{6}}$

$In the fourth quadrant, \cos x and \sec x are positive . \cos x = \frac{1}{s e c x} = \frac{1}{\sqrt{s e c^{2} x}} = \frac{1}{\sqrt{1 + \tan^{2} x}} = \frac{1}{\sqrt{1 + {(- \frac{1}{\sqrt{5}})}^{2}}} = \frac{1}{\sqrt{\frac{6}{5}}} = \frac{\sqrt{5}}{\sqrt{6}}$

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Similar questions

θ = - \frac{1}{\sqrt{5}}

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

\frac{2}{\sqrt{6}}

\frac{1}{2}

\frac{1}{\sqrt{6}}

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tan \frac{x}{2}

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