If tan x = −1/√5 and x lies in the IV quadrant, then the value of cos x is(a) √5/√6(b) 2/√6(c) 1/2(d) 1/√6

Question

\[ \text{If } \tan x=-\frac{1}{\sqrt5} \text{ and } x \text{ lies in IV quadrant,} \]

\[ \text{then the value of } \cos x \text{ is} \]

(a) \(\frac{\sqrt5}{\sqrt6}\)
(b) \(\frac{2}{\sqrt6}\)
(c) \(\frac12\)
(d) \(\frac1{\sqrt6}\)

\[ \tan x=\frac{\text{Perpendicular}}{\text{Base}} = -\frac1{\sqrt5} \]

Take

\[ \text{Perpendicular}=-1,\quad \text{Base}=\sqrt5 \]

\[ \text{Hypotenuse} = \sqrt{(-1)^2+(\sqrt5)^2} \]

\[ = \sqrt{1+5} = \sqrt6 \]

Now,

\[ \cos x=\frac{\text{Base}}{\text{Hypotenuse}} = \frac{\sqrt5}{\sqrt6} \]

Since \(x\) lies in IV quadrant, cosine is positive.

\[ \boxed{\frac{\sqrt5}{\sqrt6}} \]

Correct Option: (a)

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