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Q.

Find the value of other five trigonometric ratios, when $\sec x = \frac{13}{5}$ and $x$ lies in fourth quadrant.

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a

cosec x = - \frac{23}{12}, \cos x = \frac{1}{13}

,

\sin x = - \frac{2}{13}, \tan x = - \frac{2}{5}

,

\cot x = \frac{5}{12}

b

cosec x = \frac{5}{13}, \cos x = \frac{5}{23}, \sin x = \frac{12}{13}, tan x = \frac{12}{5}, \cot x = \frac{25}{12}

c

cosec x = \frac{13}{15}, co s x = \frac{15}{13}, \sin x = - \frac{12}{13}, tan x = - \frac{12}{51}, \cot x = \frac{45}{12}

d

cosec x = - \frac{13}{12}, co s x = \frac{5}{13}, \sin x = - \frac{12}{13}, tan x = - \frac{12}{5}, \cot x = - 5 / 12

answer is D.

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Detailed Solution

Given,

\sec x = \frac{13}{5}

Now, we know that,

\sec^{2} x = 1 + \tan^{2} x

\Rightarrow {(\frac{13}{5})}^{2} = 1 + \tan^{2} x

\Rightarrow \tan^{2} x = \frac{169}{25} - 1

\Rightarrow \tan^{2} x = \frac{144}{25}

\Rightarrow \tan x = \pm \frac{12}{5}

As,

x

lies in fourth quadrant.

∴ \tan x = \frac{- 12}{5}

Now we know that,

\cot x = \frac{1}{\tan x} = \frac{- 5}{12}

\cos x = \frac{1}{\sec x} = \frac{5}{13}

Also,

\Rightarrow \sin^{2} x + {(\frac{5}{13})}^{2} = 1 (\sin^{2} x + \cos^{2} x = 1)

\Rightarrow \sin^{2} x = 1 - \frac{25}{169}

\Rightarrow \sin^{2} x = \frac{144}{169}

\Rightarrow \sin x = \pm \frac{12}{13}

As,

x

lies in fourth quadrant.

∴ \sin x = \frac{- 12}{13}

Therefore, the results are,

cosec x = - \frac{13}{12}, \cos x = \frac{5}{13}

,

\sin x = - \frac{12}{13}, \tan x = - \frac{12}{5}

,

\cot x = \frac{- 5}{12}

Thus, option 1 is correct.

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