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}

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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

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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}

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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}}

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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}