Question 1: Find the values of the other five trigonometric functions in each of the following:

i) $\cot x =$ $\frac{12}{5}$, $x$ in Quadrant III    ii) $\cos x =$ $\frac{-1}{2}$, $x$ in Quadrant II

iii) $\tan x =$ $\frac{3}{4}$, $x$ in Quadrant III    iv) $\sin x =$ $\frac{3}{5}$, $x$ in Quadrant I

Answer:

i)   $\cot x =$ $\frac{12}{5}$, $x$ in Quadrant III

We know $\mathrm{cosec}^2 - \cot^2 x = 1 \Rightarrow \mathrm{cosec} x = \pm \sqrt{1 + \cot^2 x}$

in Quadrant III, $\mathrm{cosec} x$ is negative

$\therefore \mathrm{cosec} x = -$ $\sqrt{1 + (\frac{12}{5})^2}$ $= -$ $\sqrt{\frac{169}{25}}$ $= -$ $\frac{13}{5}$

We know, $\tan x =$ $\frac{1}{\cot x}$ $=$ $\frac{5}{12}$

Also $\sin x =$ $\frac{1}{\mathrm{cosec} x}$ $= -$ $\frac{5}{13}$

Now $\sin^2 x + \cos^2 x = 1 \Rightarrow \cos x = \pm \sqrt{1 - \sin^2 x}$

In Q III, $\cos x$ is negative

$\therefore \cos x = -$ $\sqrt{1 - (\frac{-5}{13})^2}$ $= -$ $\sqrt{\frac{144}{169}}$ $= -$ $\frac{12}{13}$

$\therefore \sec x =$ $\frac{1}{\cos x}$ $= -$ $\frac{13}{12}$

ii)   $\cos x = -$ $\frac{1}{2}$ $, x$ is in Quadrant II

We know $\sin^2 x + \cos^2 x = 1 \Rightarrow \cos x = \pm \sqrt{1 - \sin^2 x}$

In Quadrant II, $\sin x$ is positive and $\tan x$ is negative

$\therefore \sin x =$ $\sqrt{1 - (\frac{1}{2})^2 }$ $= \frac{\sqrt{3}}{2}$

$\therefore \mathrm{cosec} x =$ $\frac{1}{\sin x}$ $=$ $\frac{2}{\sqrt{3}}$

$\tan x =$ $\frac{\sin x }{\cos x}$ $=$ $\frac{\sqrt{3}}{2}$ $(-2) = - \sqrt{3}$

$\therefore \cot x =$ $\frac{1}{\tan x}$ $= -$ $\frac{1}{\sqrt{3}}$

$\sec x =$ $\frac{1}{\cos x}$ $=$ $\frac{1}{-\frac{1}{2}}$ $= -2$

iii) $\tan x =$ $\frac{3}{4}$, $x$ in Quadrant III

$\therefore \cot x =$ $\frac{1}{\tan x}$ $=$ $\frac{4}{3}$

In Quadrant III, $\mathrm{cosec} x$ is negative

$\therefore \mathrm{cosec} = - \sqrt{1 + \cot^2 x} = -$ $\sqrt{1 + (\frac{4}{3})^2}$ $= -$ $\frac{5}{3}$

$\therefore \sin x =$ $\frac{1}{\mathrm{cosec} x}$ $= -$ $\frac{3}{5}$

Also $\tan x =$ $\frac{\sin x}{\cos x}$ $\Rightarrow \cos x = -$ $\frac{3}{5}$ $.$ $\frac{4}{3}$ $= -$ $\frac{4}{5}$

$\therefore \sec x =$ $\frac{1}{\cos x}$ $= -$ $\frac{5}{4}$

iv) $\sin x =$ $\frac{3}{5}$, $x$ in Quadrant I

In Quadrant I, $\cos x$ is positive

$\therefore \cos x = \sqrt{1 - \sin^2 x} =$ $\sqrt{1 - \frac{9}{25}}$ $=$ $\frac{4}{5}$

$\therefore \tan x =$ $\frac{\sin x}{\cos x}$ $=$ $\frac{3}{5}$ $\times$ $\frac{5}{4}$ $=$ $\frac{3}{4}$

$\therefore \cot x =$ $\frac{1}{\tan x}$ $=$ $\frac{4}{3}$

and $\mathrm{cosec} =$ $\frac{1}{\sin x}$ $=$ $\frac{5}{3}$

$\\$

Question 2: If $\sin x =$ $\frac{12}{13}$ and $x$ lies in Quadrant II, find the value of $\sec x + \tan x$.

Answer:

$\sin x =$ $\frac{12}{13}$ $, x$ in Quadrant II

$\cos x$ in Quadrant II is negative

$\therefore \cos x = - \sqrt{1 - \sin^2 x} = - \sqrt{1 - \frac{144}{169} } = -$ $\frac{5}{13}$

$\therefore \sec x =$ $\frac{1}{\cos x}$ $= -$ $\frac{13}{5}$

$\tan x=$ $\frac{\sin x}{\cos x}$ $=$ $\frac{12}{13}$ $\times$ $\frac{-13}{5}$ $= -$ $\frac{12}{5}$

$\therefore \sec x + \tan x = -$ $\frac{13}{5}$ $+$ $\frac{-12}{5}$ $= -$ $\frac{25}{5}$ $= -5$

$\\$

Question 3: If $\sin x =$ $\frac{3}{5}$ and $\tan y =$ $\frac{1}{2}$ and $\frac{\pi}{2}$ $< x < \pi < y <$ $\frac{3\pi}{2}$, find the value of $8 \tan x - \sqrt{5} \sec y$.

Answer:

Given $\sin x =$ $\frac{3}{5}$ and $\tan y =$ $\frac{1}{2}$ and $\frac{\pi}{2}$ $< x < \pi < y <$ $\frac{3\pi}{2}$

$\Rightarrow x$ in Quadrant II and $y$ in Quadrant III

$\cos x$ in Quadrant II is negative

$\therefore \cos x = \sqrt{1 - \sin^2 x} = -$ $\sqrt{1 - \frac{9}{25}}$ $=$ $\frac{-4}{5}$

$\therefore \tan x=$ $\frac{\sin x}{\cos x}$ $=$ $\frac{3}{5}$ $\times$ $\frac{-5}{4}$ $= -$ $\frac{3}{4}$

$\tan y =$ $\frac{1}{2}$

$\sec^2 y = 1 + \tan^2 y = 1 + \Big($ $\frac{1}{2}$ $\Big)^2 =$ $\frac{5}{4}$

$\sec y$ in Quadrant III is negative

$\therefore \sec y = -$ $\sqrt{\frac{5}{4}}$ $= -$ $\frac{\sqrt{5}}{2}$

$\therefore 8 \tan x - \sqrt{5} \sec y = 8 \times (-$ $\frac{3}{4}$ $) - \sqrt{5} (-$ $\frac{\sqrt{5}}{2}$ $) = -6 +$ $\frac{5}{2}$ $= -$ $\frac{7}{2}$

$\\$

Question 4: If $\sin x + \cos y = 0$, and $x$ lies in the fourth quadrant, find $\sin x$ and $\cos x$.

Answer:

$\sin x + \cos x = 0, x$ in Quadrant IV

$\Rightarrow \sin x = - \cos x$

$\Rightarrow \tan x = - 1$

$\therefore \sec^2 x = 1 + \tan^2 x$

in Quadrant IV, $\sec x$ is positive

$\therefore \sec x = \sqrt{1 + (-1)^2} = \sqrt{2}$

$\therefore \cos x =$ $\frac{1}{\sec x}$ $=$ $\frac{1}{\sqrt{2}}$

Since $\tan x =$ $\frac{\sin x }{\cos x}$

$\Rightarrow \sin x = -1 \times \cos x = -1 \times$ $\frac{1}{\sqrt{2}}$ $= -$ $\frac{1}{\sqrt{2}}$

$\\$

Question 5: If $\cos x = -$ $\frac{3}{5}$ and $\pi < x <$ $\frac{3\pi}{2}$, find the value of the five trigonometric functions and hence evaluate $\frac{\mathrm{cosec} x + \cot x}{\sec x - \tan x}$.

Answer:

Given $\cos x = -$ $\frac{3}{5}$ and $\pi < x <$ $\frac{3\pi}{2}$

$\Rightarrow x$ in Quadrant III

$\sec x =$ $\frac{1}{\cos x}$ $= -$ $\frac{5}{3}$

In Quadrant III, $\sin x$ is negative

$\therefore \sin x = - \sqrt{1 - \cos^2 x} = -$ $\sqrt{1 - \frac{9}{25} }$ $= -$ $\frac{4}{5}$

$\therefore \mathrm{cosec} x=$ $\frac{1}{\sin x}$ $= -$ $\frac{5}{4}$

$\tan x =$ $\frac{\sin x }{\cos x}$ $= -$ $\frac{4}{5}$ $\times$ $\frac{-5}{3}$ $=$ $\frac{4}{3}$

$\therefore \cot x =$ $\frac{1}{\tan x}$ $=$ $\frac{3}{4}$

$\therefore$ $\frac{\mathrm{cosec} x + \cot x}{\sec x - \tan x}$ $=$ $\frac{(\frac{-5}{4}) \times \frac{3}{4}}{(\frac{-5}{3}) - \frac{4}{3}}$ $=$ $\frac{1}{6}$