Q.1: Find the values of the other five trigonometric functions if $ \cos x = -\frac{1}{2} $, $x$ lies in third quadrant.
Solution
$\cos x = -\frac{1}{2}$
$\sec x = \frac{1}{\cos x} = -2$
We know that
$$\cos^2 x + \sin^2 x = 1$$
$$\sin^2 x = 1 – \cos^2 x$$
$$\sin^2 x = 1 – \left(-\frac{1}{2}\right)^2$$
$$\sin^2 x = 1 – \frac{1}{4} = \frac{3}{4}$$
$$\sin x = \pm \frac{\sqrt{3}}{2}$$
Since $x$ lies in the third quadrant, $\sin x$ is negative.
$$\sin x = -\frac{\sqrt{3}}{2}$$
$$cosec \;x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$$
$$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$
$$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$$
Question 2
Find the values of the other five trigonometric functions if $ \sin x = \frac{3}{5} $, $x$ lies in second quadrant.
Solution
$\sin x = \frac{3}{5}$
$\csc x = \frac{1}{\sin x} = \frac{5}{3}$
We know that
$$\cos^2 x + \sin^2 x = 1$$
$$\cos^2 x = 1 – \sin^2 x$$
$$\cos^2 x = 1 – \left(\frac{3}{5}\right)^2 = 1 – \frac{9}{25} = \frac{16}{25}$$
$$\cos x = \pm \frac{4}{5}$$
Since $x$ lies in the second quadrant, $\cos x$ is negative.
$$\cos x = -\frac{4}{5}$$
$$\sec x = \frac{1}{\cos x} = -\frac{5}{4}$$
$$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}$$
$$\cot x = \frac{1}{\tan x} = -\frac{4}{3}$$
Question 3
Find the values of the other five trigonometric functions if $ \cot x = \frac{3}{4} $, $x$ lies in third quadrant.
Solution
$\cot x = \frac{3}{4}$
$\tan x = \frac{1}{\cot x} = \frac{4}{3}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\sec^2 x = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$
$$\sec x = \pm \frac{5}{3}$$
Since $x$ lies in the third quadrant, $\sec x$ is negative.
$$\sec x = -\frac{5}{3}$$
$$\cos x = \frac{1}{\sec x} = -\frac{3}{5}$$
We know that
$$1 + \cot^2 x = \csc^2 x$$
$$\csc^2 x = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}$$
$$\csc x = \pm \frac{5}{4}$$
Since $x$ lies in the third quadrant, $\csc x$ is negative.
$$\csc x = -\frac{5}{4}$$
$$\sin x = \frac{1}{\csc x} = -\frac{4}{5}$$
Question 4
Find the values of the other five trigonometric functions if $ \sec x = \frac{13}{5} $, $x$ lies in fourth quadrant.
Solution
$\sec x = \frac{13}{5}$
$\cos x = \frac{1}{\sec x} = \frac{5}{13}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\tan^2 x = \sec^2 x – 1 = \left(\frac{13}{5}\right)^2 – 1 = \frac{169}{25} – 1 = \frac{144}{25}$$
$$\tan x = \pm \frac{12}{5}$$
Since $x$ lies in the fourth quadrant, $\tan x$ is negative.
$$\tan x = -\frac{12}{5}$$
$$\cot x = \frac{1}{\tan x} = -\frac{5}{12}$$
We know that
$$\sin^2 x + \cos^2 x = 1$$
$$\sin^2 x = 1 – \cos^2 x = 1 – \left(\frac{5}{13}\right)^2 = 1 – \frac{25}{169} = \frac{144}{169}$$
$$\sin x = \pm \frac{12}{13}$$
Since $x$ lies in the fourth quadrant, $\sin x$ is negative.
$$\sin x = -\frac{12}{13}$$
$$\csc x = \frac{1}{\sin x} = -\frac{13}{12}$$
Question 5
Find the values of the other five trigonometric functions if $ \tan x = -\frac{5}{12} $, $x$ lies in second quadrant.
Solution
$\tan x = -\frac{5}{12}$
$\cot x = \frac{1}{\tan x} = -\frac{12}{5}$
We know that
$$1 + \tan^2 x = \sec^2 x$$
$$\sec^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$$
$$\sec x = \pm \frac{13}{12}$$
Since $x$ lies in the second quadrant, $\sec x$ is negative.
$$\sec x = -\frac{13}{12}$$
$$\cos x = \frac{1}{\sec x} = -\frac{12}{13}$$
We know that
$$1 + \cot^2 x = \csc^2 x$$
$$\csc^2 x = 1 + \left(-\frac{12}{5}\right)^2 = 1 + \frac{144}{25} = \frac{169}{25}$$
$$\csc x = \pm \frac{13}{5}$$
Since $x$ lies in the second quadrant, $\csc x$ is positive.
$$\csc x = \frac{13}{5}$$
$$\sin x = \frac{1}{\csc x} = \frac{5}{13}$$
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