Find the value of other five trigonometric ratios, when sec⁡x=135 and x lies in fourth quadrant.

cosec x = - 13 12 , co s ⁡ x = 5 13 , sin x = - 12 13 , tan x = - 12 5 , cot ⁡ x = - 5 / 12

Find the value of other five trigonometric ratios, when sec⁡x=135 and x lies in fourth quadrant.

cosec x = - 13 12 , co s ⁡ x = 5 13 , sin x = - 12 13 , tan x = - 12 5 , cot ⁡ x = - 5 / 12

cosec x = - 23 12 , cos ⁡ x = 1 13 , sin x = - 2 13 , tan ⁡ x = - 2 5 , cot x = 5 12

cosec x = 5 13 , cos ⁡ x = 5 23 , sin x = 12 13 , tan x = 12 5 , cot ⁡ x = 25 12

cosec x = 13 15 , co s ⁡ x = 15 13 , sin x = - 12 13 , tan x = - 12 51 , cot ⁡ x = 45 12

cosec x = - 13 12 , co s ⁡ x = 5 13 , sin x = - 12 13 , tan x = - 12 5 , cot ⁡ x = - 5 / 12

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

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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}

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}

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}

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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