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Sign of Trigonometric Ratios in Different Quadrants

If tan x=-15 ...

Question

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}}

If $θ + ϕ = α$ and $s i n θ = k s i n ϕ$ , then show that $t a n θ = \frac{k s i n α}{1 + k c o s α}$ and $t a n ϕ = \frac{s i n α}{k + c o s α}$

If $s i n x = \frac{- 1}{3}$ and x lies in quadrant IV, then find the value of

(i) $s i n \frac{x}{2}$

(ii) $c o s \frac{x}{2}$

(iii) $t a n \frac{x}{2}$

If cos x= -1/2 and x lies in quadrant II. What is the value of tan x?

tan x = \frac{3}{4}

x

I I I

sin \frac{x}{2}, cos \frac{x}{2}

tan \frac{x}{2}

tan θ = 3

θ

sin θ

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