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प्रश्न
If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is
पर्याय
- \[\frac{\sqrt{5}}{\sqrt{6}}\]
- \[\frac{2}{\sqrt{6}}\]
- \[\frac{1}{2}\]
- \[\frac{1}{\sqrt{6}}\]
उत्तर
\[\text{ In the fourth quadrant, }\cos x \text{ and }\sec x\text{ are positive . }\]
\[\cos x = \frac{1}{\sec x}\]
\[ = \frac{1}{\sqrt{\sec^2 x}}\]
\[ = \frac{1}{\sqrt{1 + \tan^2 x}}\]
\[ = \frac{1}{\sqrt{1 + \left( - \frac{1}{\sqrt{5}} \right)^2}}\]
\[ = \frac{1}{\sqrt{\frac{6}{5}}}\]
\[ = \frac{\sqrt{5}}{\sqrt{6}}\]
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